吸引子是描述无穷维维动力系统的渐近行为的重要概念,参见文献[1-17].然而,尽管吸引子具有紧性,但它可能是无穷维的.这意味着只有用无穷多个参数才能刻画无穷维动力系统的渐近行为.另一方面,吸引子吸引轨道的速度可能是任意慢的,从而吸引子在扰动下是不稳定的.为克服以上两点缺陷, Eden等^[18]提出了指数吸引子的概念,它是具有有限分形维数且以指数速率吸引轨道的紧的正不变集.用不同的方法将这一概念推广到非自治动力系统,可以得到两类指数吸引子:拉回和一致指数吸引子,参见文献[19-24].

Shirikyan等^[25]将指数吸引子推广到自治随机动力系统的随机指数吸引子并给出它的一些存在条件.与文献[25]相比较, Caraballo等^[26]减弱了文献[25]中随机指数吸引子的存在条件.最近,周盛凡等^[27-29]将指数吸引子的概念推广到连续余圈(非自治随机动力系统)的随机指数吸引子,建立了随机指数吸引子存在的判据,并应用这一判据证明了随机格点系统、无界域上的随机反应扩散方程、有界域上的随机波动方程的随机指数吸引子的存在性.相比文献[25-26]中的存在性条件,周盛凡等给出的判据更容易验证.

FitzHugh-Nagumo系统是模拟信号在轴突间传递的模型,参见文献[30-31].本文研究如下$ {{\Bbb R}} ^{3} $上的随机FitzHugh-Nagumo系统的解的渐近行为

(1.1) $ \begin{equation} \left\{ \begin{array}{l} {\rm d}\tilde{u}_{1}+(\lambda \tilde{u}_{1}-\Delta \tilde{u}_{1}+\alpha \tilde{u} _{2}){\rm d}t = (f(x, \tilde{u}_{1})+g_{1}(x, t)){\rm d}t+b\tilde{u}_{1}\circ {\rm d}W(t), \, \, \, t>\tau , \\ {\rm d}\tilde{u}_{2}+(\sigma \tilde{u}_{2}-\beta \tilde{u}_{1}){\rm d}t = g_{2}(x, t){\rm d}t+b \tilde{u}_{2}\circ {\rm d}W(t), \\ \tilde{u}_{1}(x, \tau ) = \tilde{u}_{1, \tau }(x), \; \; \tilde{u}_{2}(x, \tau ) = \tilde{u}_{2, \tau }(x), \; \; x\in {{\Bbb R}} ^{3}, \tau \in {{\Bbb R}} , \end{array} \right. \end{equation} $

其中$ \tilde{u}_{i} = \tilde{u}_{i}(x, t) $$ (i = 1, 2) $是定义在$ {{\Bbb R}} ^{3}\times \lbrack \tau , +\infty ) $上的实值函数,系数$ \lambda $, $ \alpha, $$ \sigma, $$ \beta $和$ b $是正常数. $ W(t) $是定义在概率空间$ (\Omega , {\cal F}, {{\Bbb P}}) $上的双边实值布朗运动,这里$ \Omega = \{\omega \in {C}({{\Bbb R}} , {{\Bbb R}} ): \omega (0) = 0\}, $$ {\cal F} $是由$ \Omega $上的紧开拓扑产生的Borel $ \sigma $ -代数, $ {\Bbb P} $是$ {\cal F} $上相应的Wiener测度.取布朗运动的实现为$ W(t) = W(t, \omega ) = \omega (t) $, $ \omega \in \Omega $, $ t\in {{\Bbb R}} $.符号$ "\circ " $表示随机积分是在斯特拉托诺维奇(Stratonovich)意义下.函数$ g_{1} $, $ g_{2} $, $ f $满足以下假设:

(A1) $ g_{i}\in C({{\Bbb R}} ^{3}\times {{\Bbb R}} , {{\Bbb R}} ) $, $ g_{i}(x, \cdot )\in C({{\Bbb R}} , $$ L^{2}({{\Bbb R}} ^{3})) $, $ \sup\limits_{t\in {{\Bbb R}} }\|g_{i}(\cdot , t)\|^{2}<\infty $, $ i = 1, 2 $,且对任意$ \varepsilon >0 $,存在$ R_{{i}, \varepsilon }\geq 0 $使得$ \sup\limits_{r\in {{\Bbb R}} }\int_{|x|\geq R_{{i}, \varepsilon }}g_{i}^{2}(x, r){\rm d}x\leq \varepsilon $,这里$ \Vert \cdot \Vert $表示$ L^{2}({{\Bbb R}} ^{3}) $的范数.

(A2)存在函数$ \beta _{1}\in L^{1}({{\Bbb R}} ^{3}, {{\Bbb R}} _{+}), $$ \beta _{2}, $$ \beta_{3}\in L^{2}({{\Bbb R}} ^{3}, {{\Bbb R}} _{+}), $$ \beta _{4}\in L^{3}({{\Bbb R}} ^{3}, {{\Bbb R}} _{+}) $和正常数$ c_{1}, $$ c_{2}, $$ c_{3}>0 $使得

(1.2) $ \begin{equation} \left\{ \begin{array}{l} uf(x, u)\leq \beta _{1}(x), \quad |f(x, u)|\leq c_{1}|u|^{3}+\beta _{2}(x), \\ { }\left\vert \frac{\partial f}{\partial x}(x, u)\right\vert \leq \beta _{3}(x), \quad \frac{\partial f}{\partial u}(x, u)\leq c_{2}, \\ { }\left\vert \frac{\partial f}{\partial u}(x, u)\right\vert \leq c_{3}u^{2}+\beta _{4}(x), \quad \end{array} \right. \quad u\in {{\Bbb R}} , \, \, x\in {{\Bbb R}} ^{3}. \end{equation} $

关于有界或无界域上的确定性或随机FitzHugh-Nagumo系统的各种吸引子的研究已有许多结果^[32-42].但至今尚无在无界域上的随机FitzHugh-Nagumo系统的随机指数吸引子的相关结果.

本文讨论FitzHugh-Nagumo系统(1.1)的随机指数吸引子的存在性.文献[29]的定理2.1给出了连续余圈存在随机指数吸引子的判据.注意到由于其(3.45)式中$ t\geq \frac{4}{\sigma }\ln (1+\frac{2\alpha \beta}{\sigma ^{2}}) $的限制,在证明系统(1.1)存在随机指数吸引子时,不能直接验证文献[29]中定理2.1的条件$ ({\rm H}3) $.本文修改了文献[29]中定理2.1的条件$ ({\rm H}3) $并得到可分Banach空间上的连续余圈存在随机指数吸引子的新判据(见定理2.1).根据这个新判据,论文证明了系统(1.1)存在随机指数吸引子.为此,需要对系统(1.1)的解在$ H^{1}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3}) $ (注意到系统(1.1)的第二个分量$ \tilde{u}_{2} $没有任何光滑性）中做尾部估计并在$ L^{2}({\Bbb {{\Bbb R}} }^{3})\times L^{2}({\Bbb {{\Bbb R}} }^{3}) $中将系统的两解之差分解成三部分,其中一部分属于有限维空间,另外两部分在时间变量和空间变量充分大时变得足够小.值得一提的是,本文所采用的方法不需要证明第二个分量$ \tilde{u}_{2} $有较高的正则性,这与文献[42]中有界域上的FitzHugh-Nagumo系统(1.1)存在随机吸引子的证明不同.

本文结构安排如下:第2节简要回顾随机指数吸引子的相关知识并给出随机指数吸引子存在的新判据.第3节证明FitzHugh-Nagumo系统(1.1)存在随机指数吸引子.

本节我们给出可分Banach空间上的连续余圈的随机指数吸引子存在的新判据.

设$ (X, \| X\|_{X}) $是可分Banach空间, $ {\cal B}(X) $是其Borel $ \sigma $ -代数. $ X $中任意子集$ F_{1}, F_{2} $的Hausdorff半距离定义为$ {\rm d}_{h}(F_{1}, F_{2}) = \sup\limits_{u\in F_{1}}\inf\limits_{v\in F_{2}}\|u-v\|_{X} $.令$ (\Omega , {\cal F}, {\Bbb P}, \{\theta _{t}\omega \}_{t\in {{\Bbb R}} }) $是遍历度量动力系统^[9].随机变量$ \gamma _{\omega } $称为关于$ \{\theta _{t}\omega \}_{t\in {{\Bbb R}} } $是缓增的,如果对于a.e. $ \omega \in \Omega $,有$ \limsup\limits_{t\rightarrow \infty }e^{-\epsilon |t|}|\gamma _{\theta_{t}\omega }| = 0 $ ($ \forall $$ \epsilon >0) $,参见文献[9].

定义2.1^[12]  称$ X $的一族非空子集$ D = \{D(\tau, \omega )\subset X:\tau \in {{\Bbb R}} , \omega \in \Omega \} $关于$ \{\theta _{t}\omega\}_{t\in {{\Bbb R}} } $是缓增的,如果对任意的$ \tau \in {{\Bbb R}} $及a.e. $ \omega \in \Omega $, $ \lim\limits_{t\rightarrow -\infty }e^{\epsilon t}\| D(\tau +t, \theta _{t}\omega )\|_{X} = 0 $, $ \forall\epsilon >0 $,其中$ \|D(\tau , \omega )\|_{X} = \sup\limits_{x\in D(\tau , \omega )}\|x\|_{X} $.

用$ {\cal D}(X) $表示$ X $的所有缓增集族构成的集合.

定义2.2^[12]  称映射$ \Psi :{{\Bbb R}} ^{+}\times{{\Bbb R}} \times\Omega \times X\rightarrow X $为$ {{\Bbb R}} $和$ (\Omega , {\cal F}, {\Bbb P}, \{\theta _{t}\omega \}_{t\in {{\Bbb R}} }) $驱动的空间$ X $上的连续余圈,如果: (ⅰ) $ \Psi (\cdot, \tau , \cdot , \cdot ) $: $ {{\Bbb R}} ^{+}\times \Omega \times X\rightarrow X $是$ ({\cal B}({{\Bbb R}} ^{+})\times {\cal F}\times {\cal B}(X), {\cal B}(X)) $ -可测的; (ⅱ) $ \Psi (0, \tau , \omega , \cdot ) $是$ X $上的恒等映射; (ⅲ) $ \Psi (t+s, \tau , \omega , \cdot ) = \Psi (t, \tau +s, \theta _{s}\omega , \Psi (s, \tau , \omega , \cdot )) $, $ \forall t, s\geq 0 $, $ \tau\in {{\Bbb R}} $; (ⅳ) $ \Psi (t, \tau , \omega , \cdot ) $: $ X\rightarrow X $是连续的.

定义2.3^[27]  称$ X $的非空子集族$ \{{\cal E}(\tau , \omega )\}_{\tau \in {{\Bbb R}} , \omega \in \Omega } $是连续余圈$ \{\Psi (t, \tau , \omega )\}_{t\geq 0, \tau \in {{\Bbb R}} , \omega \in \Omega } $的$ {\cal D}(X) $ -随机指数吸引子,如果存在全测集$ \tilde{\Omega}\in {\cal F} $使得对任意$ \tau \in {{\Bbb R}} $及$ \omega \in \tilde{\Omega} $,满足: (ⅰ) $ {\cal E}(\tau , \omega ) $是$ X $中的紧子集且关于$ \omega $可测; (ⅱ)存在随机变量$ \varsigma _{\omega } $$ (<\infty ) $使得分形维数$ \sup\limits_{\tau \in {{\Bbb R}} }\dim _{f} {\cal E}(\tau , \omega )\leq \varsigma _{\omega } $,其中$ \dim _{f}{\cal E}(\tau , \omega ) = \limsup\limits_{\varepsilon \rightarrow 0+}\frac{\ln N_{\varepsilon }({\cal E}(\tau , \omega ))}{-\ln \varepsilon } $, $ N_{\varepsilon }({\cal E}(\tau , \omega )) $是$ X $中能覆盖$ {\cal E}(\tau , \omega ) $所需的半径为$ \varepsilon $的球的最小个数; (ⅲ) $ \Psi (t, \tau -t, \theta _{-t}\omega ){\cal E}(\tau -t, \theta _{-t}\omega )\subseteq {\cal E}(\tau, \omega ) $, $ \forall t\geq0 $；(ⅳ)存在常数$ \tilde{a}>0 $,使得对任意$ B\in {\cal D}(X) $,存在随机变量$ t_{B}(\tau , \omega )\geq 0 $和$ Q(\tau , \omega , \|B\|_{X})>0 $满足

$ {\rm d}_{h}(\Psi (t, \tau -t, \theta _{-t}\omega )B(\tau -t, \theta _{-t}\omega ), {\cal E}(\tau , \omega ))\leq Q(\tau , \omega , \|B\|_{X})e^{-\tilde{a}t}, \quad t\geq t_{B}(\tau , \omega ). $

在下文中,为方便起见,将" a.e. $ \omega \in \Omega $"写为"$ \omega \in \Omega $".基于文献[29]中定理2.1的证明,将该定理证明稍作修正和改进可得到如下定理:

定理2.1  令$ \{\Psi (t, \tau , \omega)\}_{t\geq 0, \tau \in {{\Bbb R}} , \omega \in \Omega } $是$ {{\Bbb R}} $和$ (\Omega , {\cal F}, {\Bbb P}, \{\theta _{t}\omega \}_{t\in {{\Bbb R}} }) $驱动的可分Banach空间$ X $上的连续余圈.假设

(H1)   存在$ X $的缓增的随机闭子集族$ \{\check{\chi}(\tau , \omega )\}_{\tau \in {{\Bbb R}} , \omega \in \Omega } $使得对任意$ \tau \in {{\Bbb R}} $及$ \omega \in \Omega $,

(h11)   $ \check{\chi}(\tau , \omega) $的直径有界,即$ \sup\limits_{\tau \in {{\Bbb R}} }\sup\limits_{u, v\in \check{\chi}(\tau , \omega)}\|u-v\|_{X}\leq R_{\omega }<\infty , $其中$ R_{\omega } $ (独立于$ \tau $)是缓增随机变量且$ R_{\theta_{t}\omega } $关于$ t\in {{\Bbb R}} $连续;

(h12)   $ \check{\chi}(\tau, \omega ) $是正不变的,即$ \Psi (t, \tau -t, \theta _{-t}\omega )\check{\chi }(\tau -t, \theta _{-t}\omega )\subseteq \check{\chi}(\tau , \omega ) $, $ \forall t\geq 0 $;

(H2)   存在正常数$ \hat{\lambda} $, $ \hat{\delta} $, $ \hat{t}_{0} $,随机变量$ \hat{C}_{0}(\omega) $, $ \hat{C}_{1}(\omega) \geq 0 $及$ N $ -维投影$ P_{N} $: $ X\rightarrow P_{N}X $ {\rm($ \dim (P_{N}X) = N\in{\Bbb N} $)},使得对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega $及$ u, v\in\check{\chi}(\tau, \omega) $,有

$ \|\Psi (t, \tau , \omega )u-\Psi (t, \tau , \omega )v\|_{X}\leq e^{\int_{0}^{ \hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s}\|u-v\|_{X}, \quad \forall t\in \lbrack 0, \hat{t}_{0}], $

$ \|(I-P_{N})\left( \Psi (t_{0}, \tau , \omega )u-\Psi (t_{0}, \tau , \omega )v\right)\|_{X} \leq(e^{-\hat{\lambda}\hat{t}_{0}+\int_{0}^{\hat{t}_{0}}\hat{C} _{1}(\theta _{s}\omega ){\rm d}s}+\frac{\hat{\delta}}{2}e^{\int_{0}^{\hat{t}_{0}} \hat{C}_{0}(\theta _{s}\omega ){\rm d}s})\|u-v\|_{X}, $

其中$ \hat{\lambda} $, $ \hat{\delta} $, $ \hat{t}_{0} $, $ N $与$ \tau $和$ \omega $无关;

(H3) $ \hat{C}_{0}(\omega ), $$ \hat{C}_{1}(\omega ), $$ \hat{\lambda} $, $ \hat{t}_{0}, $$ \hat{\delta} $满足:

$ \begin{eqnarray*} \left\{ \begin{array}{l} { } \hat{t}_{0}\geq \frac{4\ln 2}{\hat{\lambda}}>0, \qquad 0\leq {\bf E[}\hat{C }_{1}(\omega )]\leq \frac{\hat{\lambda}}{8}, \quad 0\leq {\bf E}[\hat{C} _{0}^{2}(\omega )]<\infty , \\ { } 0<\hat{\delta}\leq \min \Big\{ \frac{1}{4}e^{-\frac{\hat{\lambda}\hat{t}_{0}}{2}}, e^{-\frac{\hat{t}_{0}^{2}}{\ln \frac{3}{2}}\left( {\bf E[} C_{0}^{2}(\omega)]+\hat{\lambda}{\bf E[}C_{0}(\omega )]\right) }\Big\}, \end{array} \right. \quad \label{8.5} \end{eqnarray*} $

其中"$ {\bf E} $"表示期望.

那么$ \{\Psi (t, \tau , \omega )\}_{t\geq 0, \tau \in {{\Bbb R}} , \omega \in \Omega } $存在一个随机指数吸引子$ \{{\cal E}(\tau , \omega )\}_{\tau \in {{\Bbb R}} , \omega \in \Omega} $,且具有以下性质:对任意$ \tau \in {{\Bbb R}} $和$ \omega \in\Omega $,

(ⅰ) $ {\cal E}(\tau, \omega ) $$ (\subseteq \check{\chi}(\tau, \omega)) $是$ X $中的紧集,且关于$ \omega $可测;

(ⅱ) $ \Psi (t, \tau -t, \theta _{-t}\omega ){\cal E}(\tau -t, \theta _{-t}\omega )\subseteq {\cal E}(\tau , \omega ) $, $ \forall t\geq 0 $;

(ⅲ) $ \dim _{f}{\cal E}(\tau , \omega )\leq \frac{2N\ln \left(\frac{2\sqrt {N}}{\hat{\delta}}+1\right)}{\ln \frac{4}{3}}<\infty $;

(ⅳ)存在随机变量$ \tilde{T}_{\omega }\geq 0 $和缓增随机变量$ \breve{b}_{\omega }>0 $,使得

$ \begin{eqnarray*} {\rm d}_{h}\left( \Psi (t, \tau , \omega )\check{\chi}(\tau , \omega ), {\cal E}(t+\tau , \theta _{t}\omega )\right) \leq \breve{b}_{\omega }e^{- \frac{\ln \frac{4}{3}}{4\hat{t}_{0}}t}, \; \; t\geq \tilde{T}_{\omega}. \label{8.5-1} \end{eqnarray*} $

证  注意到该定理的条件与文献[29]中定理2.1的条件中唯一不同的是关于$ \hat{C}_{0}(\omega ), $$ \hat{C}_{1}(\omega ) $, $ \hat{ \lambda} $, $ \hat{t}_{0} $, $ \hat{\delta} $的条件(H3).条件(H3)来自以下要求:对任意固定$ \omega \in \Omega $和$ m\in{\Bbb Z} $,存在整数$ n_{0}(\omega )\in {\Bbb N} $使得对任意$ n\geq n_{0}(\omega ) $, $ \prod\limits_{l = 1}^{n}a_{ \theta _{(m-l)\hat{t}_{0}}\omega } $满足

(2.1) $ \begin{equation} \prod\limits_{l = 1}^{n}a_{\theta _{(m-l)\hat{t}_{0}}\omega }\leq\left( \frac{3}{4} \right) ^{n}, \end{equation} $

其中

$ a_{\theta _{(m-l)\hat{t}_{0}}\omega } = 2e^{-\hat{\lambda}\hat{t} _{0}+\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s}+2\hat{\delta}e^{\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s}, \quad l = 1, \cdots, n. \label{8.5-3} $

现在证明条件(H3)能推出(2.1)式,这其实是对定理2.1^[29]的证明中的相应部分的改进.令

(2.2) $ \begin{equation} J = \left\{ \omega \in \Omega :\int_{0}^{\hat{t}_{0}}[\hat{C}_{0}(\theta _{s}\omega )-\hat{C}_{1}(\theta _{s}\omega )]{\rm d}s+\hat{\lambda}\hat{t}_{0}>\ln \frac{1}{\hat{\delta}}\right\} . \end{equation} $

(a)若$ \theta _{(m-l)\hat{t}_{0}}\omega \in J, $那么

$ \hat{\delta}e^{\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s}>e^{-\hat{\lambda}\hat{t}_{0}+\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s}, $

$ a_{\theta _{(m-l)\hat{t}_{0}}\omega }<4\hat{\delta}e^{\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s}, \quad l\geq 1. $

(b)若$ \theta _{(m-l)\hat{t}_{0}}\omega \notin J, $那么

$ \hat{\delta}e^{\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s}\leq e^{-\hat{\lambda}\hat{t}_{0}+\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s}, $

$ a_{\theta _{(m-l)\hat{t}_{0}}\omega }\leq 4e^{-\hat{\lambda}\hat{t} _{0}+\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s}, l\geq 1. $

记

$ \sum\limits_{l = 1}^{n}\vartheta _{J}(\theta _{(m-l)\hat{t}_{0}}\omega ) = k_{1}, \quad \sum\limits_{l = 1}^{n}[1-\vartheta _{J}(\theta _{(m-l)\hat{t}_{0}}\omega )] = k_{2}, \quad n = k_{1}+k_{2}. $

根据(H3),可得$ 2e^{-\frac{\hat{\lambda}\hat{t}_{0}}{2}}\leq \frac{1}{ 2} $, $ 8\hat{\delta}e^{\frac{\hat{\lambda}\hat{t}_{0}}{2}}\leq 2 $,且

(2.3) $ \begin{eqnarray} \prod\limits_{l = 1}^{n}a_{\theta _{(m-l)\hat{t}_{0}}\omega } &\leq& \left( 4\hat{\delta}\right) ^{^{k_{1}}}e^{\sum\limits_{l = 1}^{n}\vartheta _{J}(\theta _{(m-l)\hat{t}_{0}}\omega )\cdot \int_{(m-l)\hat{t}_{0}}^{(m-l+1) \hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s} {} \\ &&\times 4^{k_{2}}e^{\sum\limits_{l = 1}^{n}[1-\vartheta _{J}(\theta _{(m-l) \hat{t}_{0}}\omega )]\left( -\hat{\lambda}\hat{t}_{0}+\int_{(m-l)\hat{t} _{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s\right) } {} \\ &\leq &\frac{1}{2^{k_{1}}}\left( 2e^{-\frac{\hat{\lambda}\hat{t}_{0}}{2} }\right) ^{^{k_{2}}}e^{\sum\limits_{l = 1}^{n}\vartheta _{J}(\theta _{(m-l)\hat{t} _{0}}\omega )\cdot \int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C} _{0}(\theta _{s}\omega ){\rm d}s} {} \\ &&\times 2^{k_{2}}\left( 8\hat{\delta}e^{\frac{\hat{\lambda}\hat{t}_{0} }{2}}\right) ^{^{k_{1}}}e^{-\frac{\hat{\lambda}\hat{t}_{0}}{2}k_{1}-\frac{ \hat{\lambda}\hat{t}_{0}}{2}k_{2}+\sum\limits_{l = 1}^{n}[1-\vartheta _{J}(\theta _{(m-l)\hat{t}_{0}}\omega )]\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{ C}_{1}(\theta _{s}\omega ){\rm d}s} {} \\ &\leq &\left( \frac{1}{2}\right) ^{^{n}} e^{\sum\limits_{l = 1}^{n}\vartheta _{J}(\theta _{(m-l)\hat{t}_{0}}\omega )\cdot \int_{(m-l)\hat{t}_{0}}^{(m-l+1) \hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s}{}\\ &&\times 2^{n}e^{-\frac{\hat{ \lambda}\hat{t}_{0}}{2}n+\sum\limits_{l = 1}^{n} \int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t }_{0}} \hat{C}_{1}(\theta _{s}\omega ){\rm d}s}, \end{eqnarray} $

其中$ \vartheta _{J}(\omega ) = \left\{ \begin{array}{ll} 1, & \omega \in J, \\ 0, & \omega \notin J. \end{array} \right. $由Birkhoff遍历定理^[9]可得$ {\bf E[}\hat{C}_{i}^{2}(\theta _{s}\omega )] = {\bf E[}\hat{C}_{i}^{2}(\omega )] $, $ s\in {{\Bbb R}} $, $ i = 0, 1 $,且

$ \frac{1}{n}\sum\limits_{l = 1}^{n}\vartheta _{J}(\theta _{(m-l)\hat{t}_{0}}\omega )\cdot \int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s \stackrel{n\rightarrow +\infty }{\longrightarrow }{\bf E}\left( \vartheta _{J}(\omega )\int_{0}^{\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\right) , $

$ \frac{1}{n}\sum\limits_{l = 1}^{n}\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s \stackrel{n\rightarrow +\infty }{\longrightarrow} {\bf E}\left( \int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s\right) = \hat{t}_{0}{\bf E}[\hat{C}_{1}(\omega )]. $

从以上两式知,对于任意$ \omega \in \Omega $,存在整数$ n_{0}(\omega )\in {\Bbb N} $使得,对任意$ n\geq n_{0}(\omega) $有

$ \sum\limits_{l = 1}^{n}\vartheta _{J}(\theta _{(m-l)\hat{t}_{0}}\omega )\cdot \int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\leq 2n{\bf E}\left( \vartheta _{J}(\omega )\cdot \int_{0}^{\hat{t} _{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\right), $

$ \sum\limits_{l = 1}^{n}\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s\leq 2\hat{t}_{0}n{\bf E}[\hat{C}_{1}(\omega )]\leq \frac{\hat{\lambda}\hat{t}_{0}}{4}n. $

根据(2.2)式和Hölder不等式可知,对$ n\geq n_{0}(\omega ) $有

$ \begin{eqnarray*} &&{\bf E}\left( \vartheta _{J}(\omega )\cdot \int_{0}^{\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\right) \\ & = &{\bf E}\left( \frac{\zeta _{J}(\omega )\cdot \int_{0}^{\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\left( \int_{0}^{\hat{t}_{0}}[\hat{C}_{0}(\theta _{s}\omega )-\hat{C}_{1}(\theta _{s}\omega )]{\rm d}s+\hat{\lambda}\hat{t}_{0}\right) }{\int_{0}^{\hat{t}_{0}}[\hat{C}_{0}(\theta _{s}\omega )-\hat{C}_{1}(\theta _{s}\omega )]{\rm d}s+\hat{\lambda}\hat{t}_{0}}\right) \\ &\leq &{\bf E}\left( \frac{\zeta _{J}(\omega )\left[ \left( \int_{0}^{\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\right) ^{2}+\hat{\lambda}\hat{t}_{0}\int_{0}^{\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\right] }{\ln \frac{1}{\hat{\delta}}}\right) \ \\ &\leq &\frac{1}{\ln \frac{1}{\hat{\delta}}}{\bf E}\left( \hat{t} _{0}\int_{0}^{\hat{t}_{0}}C_{0}^{2}(\theta _{s}\omega ){\rm d}s+\hat{\lambda}\hat{t }_{0}\int_{0}^{\hat{t}_{0}}\hat{C}_{0}(\theta _{s}\omega ){\rm d}s\right) \\ \ &\leq &\frac{\hat{t}_{0}^{2}}{\ln \frac{1}{\hat{\delta}}}\left( {\bf E[}\hat{C}_{0}^{2}(\omega ))]+\hat{\lambda}{\bf E[}\hat{C}_{0}(\omega )]\right) , \end{eqnarray*} $

由(H3)可以推出, $ 2e^{-\frac{\hat{\lambda}\hat{t}_{0}}{4}}\leq 1 $,因此

$ \begin{eqnarray*} 2^{n}e^{-\frac{\hat{\lambda}\hat{t}_{0}}{2}n+\sum\limits_{l = 1}^{n}\int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C}_{1}(\theta _{s}\omega ){\rm d}s}\leq 2^{n}e^{-\frac{\hat{\lambda}\hat{t}_{0}}{2}n+\frac{\hat{\lambda}\hat{t}_{0}}{4}n} = \left( 2e^{-\frac{\hat{\lambda}\hat{t}_{0}}{4}}\right) ^{n}\leq 1. \end{eqnarray*} $

根据(2.3)式可以得到

$ \begin{eqnarray*} \prod\limits_{l = 1}^{n}a_{\theta _{(m-l)\hat{t}_{0}}\omega }& \leq &\left(\frac{1}{2 }\right) ^{^{n}}e^{\sum\limits_{l = 1}^{n}\vartheta _{J}(\theta _{(m-l)\hat{t} _{0}}\omega )\cdot \int_{(m-l)\hat{t}_{0}}^{(m-l+1)\hat{t}_{0}}\hat{C} _{0}(\theta _{s}\omega ){\rm d}s} \\ &\leq & \left( \frac{1}{2}e^{\frac{\hat{t}_{0}^{2}}{\ln \frac{1}{\hat{\delta}} }\left( {\bf E[}\hat{C}_{0}^{2}(\omega ))]+\hat{\lambda}{\bf E[}\hat{C} _{0}(\omega )]\right) }\right) ^{n} \leq \left( \frac{3}{4}\right) ^{n}, \quad n\geq n_{0}(\omega ). \end{eqnarray*} $

证明的其余部分类似于文献[29,定理2.1]中相应的部分.

在这一节中,考虑系统(1.1)的解产生的连续余圈的随机指数吸引子的存在性.这里我们总是假设(1.1)中的函数$ f, $$ g_{1}, $$ g_{2} $满足条件(A1)–(A2),概率空间$ (\Omega , {\cal F}, {\Bbb P}) $如引言中所定义.

令$ (\cdot , \cdot ) $, $ \|\cdot\| $和$ (\cdot , \cdot )_{1} $, $ \|\cdot\|_{1} $分别表示$ L^{2}({{\Bbb R}} ^{3}) $和$ H^{1}({{\Bbb R}} ^{3}) $的内积和诱导范数,其中

$ (v_{1}, v_{2}) = \int_{{{\Bbb R}} ^{3}}v_{1}(x)v_{2}(x){\rm d}x, \quad \, (v_{1}, v_{2})_{1} = (\nabla v_{1}, \nabla v_{2})+(v_{1}, v_{2}). $

令$ E = L^{2}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3}) $和$ E_{1} = H^{1}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3}) $是分别具有如下内积的Hilbert空间

$ (u_{1}, u_{2})_{E} = \beta (u_{11}, u_{21})+\alpha (u_{12}, u_{22}), \quad \forall u_{i} = (u_{i1}, u_{i2})\in E, \quad i = 1, 2, $

$ (u_{1}, u_{2})_{E_{1}} = (\nabla u_{11}, \nabla u_{21})+\beta (u_{11}, u_{21})+\alpha (u_{12}, u_{22}), \quad \forall u_{i} = (u_{i1}, u_{i2})\in E_{1}, \quad i = 1, 2, $

$ \| \cdot \|_{E} $和$ \|\cdot \| _{E_{1}} $表示相应的诱导范数.由嵌入$ H^{1}({{\Bbb R}} ^{3})\hookrightarrow L^{6}({{\Bbb R}} ^{3}) $^[43]知,存在$ C_{0}>0 $使得

(3.1) $ \begin{equation} \|u\|_{L^{6}({{\Bbb R}} ^{3})}\leq C_{0}\|u\|_{1} = C_{0}(\|\nabla u\|^{2}+\|u\|^{2})^{\frac{1}{2}}, \quad \forall u\in H^{1}({{\Bbb R}} ^{3}). \end{equation} $

定义样本空间$ \Omega $上的映射群$ \{\theta _{t}\}_{t\in {{\Bbb R}} } $,其中$ \theta _{t}\omega (\cdot ) = \omega (t+\cdot )-\omega (t) $, $ \omega \in \Omega $, $ t\in {{\Bbb R}} $,那么$ (\Omega , {\cal F}, {\Bbb P}, \{\theta _{t}\}_{t\in {{\Bbb R}} }) $是遍历度量动力系统.注意到$ z(\theta_{t}\omega ) = -\int_{-\infty }^{0}e^{s}(\theta _{t}\omega )(s){\rm d}s $$ (t\in {{\Bbb R}} ) $是方程d$ z+z $d$ t = {\rm d}W(t) $的平稳解,且对任意$ \omega \in \Omega $, $ z(\theta _{t}\omega ) $关于$ t $连续,此外$ \lim\limits_{t\rightarrow \pm \infty }\frac{|z(\theta _{t}\omega )|}{|t|} = \lim\limits_{t\rightarrow \pm \infty }\frac{1}{t}\int_{0}^{t}z(\theta _{s}\omega ){\rm d}s = 0 $,参见文献[14, 44].引进变量变换$ u_{i}(t, \omega, x) = e^{-bz(\theta _{t}\omega )}\tilde{u}_{i}(t, x), $$ i = 1 $, $ 2 $,那么系统(1.1)等价于下面的随机系统:

(3.2) $ \begin{equation} \left\{ \begin{array}{l} { } \frac{{\rm d}u_{1}}{{\rm d}t}-\Delta u_{1}+\lambda u_{1}-bz(\theta _{t}\omega )u_{1}+\alpha u_{2} = e^{-bz(\theta _{t}\omega )}(f(x, e^{bz(\theta _{t}\omega )}u_{1})+g_{1}(x, t)), \\ { }\frac{{\rm d}u_{2}}{{\rm d}t}+\sigma u_{2}-bz(\theta _{t}\omega )u_{2}-\beta u_{1} = e^{-bz(\theta _{t}\omega )}g_{2}(x, t), \quad t>\tau , \\ u_{i}(\tau , \omega , x) = e^{-bz(\theta _{\tau }\omega )}\tilde{u}_{i, \tau }(x) = u_{i, \tau }(\omega , x), \quad x\in {{\Bbb R}} ^{3}, \tau \in {{\Bbb R}} , i = 1, 2. \end{array} \right. \end{equation} $

由文献[34-35, 38]知,对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega $和$ \psi _{\tau }(\omega ) = (u_{1, \tau }(\omega, x), u_{2, \tau }(\omega , x))\in E $,系统(3.2)存在惟一的全局解$ \psi (t, \tau , \omega , \psi _{\tau }) = (u_{1}(t, \tau , \omega , \psi _{\tau }), u_{2}(t, \tau , \omega, \psi _{\tau })) $, $ t\in \lbrack \tau , +\infty ) $,且$ \psi (\cdot , \tau , \omega , \varphi _{\tau })\in C([\tau, +\infty );E)\cap L_{loc}^{2}([\tau , +\infty );E_{1}); $此外, $ \psi(t, \tau , \omega , \psi _{\tau }) $关于$ \omega $可测,关于初值$ \varphi _{\tau } $连续.因此,解映射簇生成一个由$ {{\Bbb R}} $和$ (\Omega , {\cal F}, {\Bbb P}, (\theta _{t})_{t\in{{\Bbb R}} }) $驱动的连续余圈$ \Psi :{{\Bbb R}} ^{+}\times {{\Bbb R}} \times \Omega \times E\rightarrow E $,

$ \begin{eqnarray*} \Psi (t, \tau , \omega , \psi _{\tau }(\omega )) & = &\Psi (t, \tau , \omega )\psi_{\tau }(\omega ) = \psi (t+\tau , \tau , \theta _{-\tau }\omega , \psi _{\tau }(\theta _{-\tau }\omega )) \\ & = &(u_{1}(t+\tau , \tau , \theta _{-\tau }\omega , \psi _{\tau } (\theta_{-\tau }\omega )), u_{2}(t+\tau , \tau , \theta _{-\tau }\omega , \psi _{\tau }(\theta _{-\tau }\omega ))), \end{eqnarray*} $

其中$ \Psi (0, \tau , \omega )\psi _{\tau }(\omega ) = \psi_{\tau }(\theta _{-\tau }\omega ) $.

对任意$ \tau \in {{\Bbb R}} $, $ \omega\in\Omega $和$ t\geq 0 $,设$ \psi (r) = \psi (r, \tau -t, \theta _{-\tau }\omega , \psi _{\tau-t}(\theta _{-\tau }\omega )) = (u_{1}(r), u_{2}(r))\in E $ ($ r\geq \tau -t $)是系统(3.2)的具有初值$ \psi (\tau -t) = \psi _{\tau-t}(\theta _{-\tau }\omega )\in E $的解.用$ {\cal D} = {\cal D}(E) $表示$ E $的所有缓增集族$ D = \{D(\tau , \omega)\subseteq E:\tau \in {{\Bbb R}} , \omega \in \Omega \} $所构成的集合.

对任意$ \tau \in {{\Bbb R}}, $$ \omega \in \Omega $和$ D\in {\cal D}(E), $令

$ {\rho } = \min \{\lambda , \frac{\sigma }{2}\}, \quad {c}_{4} = 2\beta \|\beta_{1}\|_{L^{1}}+\frac{\beta }{\lambda }\|g_{1}\|^{2}+\frac{\alpha }{\sigma }\| g_{2}\|^{2}, $

(3.3) $ \begin{equation} L_{0}^{2}(\omega ) = 1+{c}_{4}Q_{0}(\omega ), \quad Q_{0}(\omega) = \int_{-\infty }^{0}e^{\, \frac{{\rho }}{2}s-2bz(\theta _{s}\omega)+\int_{s}^{0}2bz(\theta _{l}\omega )){\rm d}l}{\rm d}s, \end{equation} $

$ {T}_{D}(\tau , \omega ) = \min \Big\{ t:e^{\int_{-t}^{0}2bz(\theta _{s}\omega ){\rm d}s-\frac{{\rho }}{2}t}\sup\limits_{\psi \in D(\tau -t, \theta _{-t}\omega )}\|\psi\|_{E}^{2}\leq 1\Big\}, $

$ D_{0}(\omega ) = \{\psi \in E:\|\psi\|_{E}\leq L_{0}(\omega )\}\subset E, $

$ {T}_{D_{0}}(\omega ) = \min \Big\{ t:e^{\int_{-t}^{0}2bz(\theta _{s}\omega ){\rm d}s-\frac{{\rho }}{2}t}\sup\limits_{\psi \in D_{0}(\theta _{-t}\omega )}\|\psi\|_{E}^{2}\leq 1\Big\}, $

$ {T}_{1, D_{0}}(\omega ) = 1+\min \left\{ t:L_{0}^{2}(\theta _{-t}\omega )e^{\frac{{\rho }}{2}(1-t)+\int_{-1}^{0}2b|z(\theta _{l}\omega )|{\rm d}l+\int_{-t}^{0}2bz(\theta _{l}\omega ){\rm d}l}\leq 1\right\} , $

(3.4) $ \begin{equation} L_{1}^{2}(\omega ) = {c}_{5}\int_{-1}^{\, 0}e^{-2bz(\theta _{s}\omega )}{\rm d}s+ \left[ \frac{1}{\beta }\left( 1+c_{2}+b\max\limits_{-1\leq s\leq 0}|z(\theta _{s}\omega )|\right) +\frac{4\alpha }{\sigma }\right] \tilde{Q} _{1}^{2}(\omega ), \end{equation} $

其中

(3.5) $ \begin{equation} \tilde{Q}_{1}^{2}(\omega ) = 1+{c}_{4}Q_{0}(\omega )e^{\, \frac{{\rho }}{2}+\int_{-1}^{0}2b|z(\theta _{l}\omega )|{\rm d}l}, \quad {c}_{5} = \|\beta_{3}\|^{2}+2\|g_{1}\|^{2}, \end{equation} $

(3.6) $ \begin{equation} L_2^2(\omega) = L_0^2(\omega)+L_1^2(\omega), \end{equation} $

$ D_{1}(\omega ) = \{\psi \in E_{1}:\|\psi\|_{E_{1}}\leq L_{2}(\omega)\}\subset E_{1}. $

关于系统(3.2)的解有如下结果.

引理3.1  对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega, $

(ⅰ)系统(3.2)的具有初值$ \psi_{\tau-t}(\theta_{-\tau}\omega)\in D(\tau -t, \theta _{-t}\omega ) \in D\in {\cal D}(E) $的解$ \psi(r) = (u_{1}(r), u_{2}(r)) $满足

$ \begin{eqnarray*} &&\|\psi (\tau , \tau - t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta_{-\tau }\omega))\|_{E}^{2} + \int_{\tau -t}^{\tau }e^{\int_{s}^{\tau}(2bz(\theta _{l-\tau }\omega )-\frac{\rho }{2}){\rm d}l}(2\beta \|\nabla u_{1}(s)\|^{2} + \frac{\sigma \alpha }{2}\| u_{2}(s)\|^{2}){\rm d}s \\ &\leq &L_{0}^{2}(\omega ), \, \, \, \, \, \, \, {\ }\forall t\geq T_{D}(\tau , \omega) \end{eqnarray*} $

和

$ \Psi (t, \tau -t, \theta _{-t}\omega )D_{0}(\theta _{-t}\omega )\, {\subseteq }\, D_{0}(\omega )\subseteq L^{2}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3}), \quad \forall t\geq {T}_{D_{0}}(\omega ); \label{4.7-0} $

(ⅱ)系统(3.2)的具有初值$ \psi_{\tau-t}(\theta_{-\tau}\omega)\in D_0(\theta_{-t}\omega ) $的解$ \psi(r) = (u_{1}(r), u_{2}(r)) $满足

$ \|\nabla u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta_{-\tau }\omega ))\| \leq {L}_{1}(\omega ), \, \, \, t\geq {T}_{1, D_{0}}(\omega ); $

(ⅲ)系统(3.2)的具有初值$ \psi_{\tau-t}(\theta_{-\tau}\omega)\in D_0(\theta_{-t}\omega ) $的解$ \psi(r) = (u_{1}(r), u_{2}(r)) $满足

$ \begin{eqnarray*} &&\Vert \psi (\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))\Vert _{E_{1}}\\ & = &\|\nabla u_{1}(\tau , \tau-t, \theta _{-\tau }\omega , \psi _{\tau-t}(\theta _{-\tau }\omega ))\|^{2}+\beta \|u_{1}(\tau , \tau -t, \theta_{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))\|^{2} \\ &&+\alpha \|u_{2}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau-t}(\theta _{-\tau }\omega))\|^{2} \\ &\leq &{L}_{2}^{2}(\omega), \quad t\geq {T}_{1, D_{0}}(\omega )+{T}_{D_{0}}(\omega), \end{eqnarray*} $

即

$ \Psi(t, \tau-t, \theta_{-t}\omega)D_0(\theta _{-t}\omega) \subseteq D_1(\omega), \quad t\geq {T}_{1, D_{0}}(\omega )+{T}_{D_{0}}(\omega). $

证  (ⅰ)取内积$ \big((3.2)_1 $, $ \beta u_{1}\big) $和$ \big((3.2)_2 $, $ \alpha u_{2}\big) $并相加,再由以下估计

$ 2\beta e^{-bz(\theta_{r-\tau }\omega )}\int_{{{\Bbb R}} ^3} f(x, e^{bz(\theta_{r-\tau }\omega )}u_1)u_1{\rm d}x\leq2\beta e^{-2bz(\theta_{r-\tau }\omega )}\|\beta_1\|_{L^1}, $

$ 2e^{-bz(\theta_{r-\tau }\omega )}(g_1, \beta u_1)\leq \frac{\beta}{\lambda}e^{-2bz(\theta_{r-\tau }\omega )}\|g_1\|^2+\lambda\beta\|u_1\|^2, $

$ 2e^{-bz(\theta_{r-\tau }\omega )}(g_2, \alpha u_2)\leq \frac{\alpha}{\sigma}e^{-2bz(\theta_{r-\tau}\omega )}\|g_2\|^2+\sigma\alpha\|u_2\|^2, $

可得

(3.7) $ \begin{equation} \frac{\rm d}{{\rm d}t}\Vert \psi \Vert _{E}^{2}+\left( \frac{\rho }{2}-2bz(\theta_{r-\tau }\omega )\right) \Vert \psi \Vert _{E}^{2}+2\beta \Vert \nabla u_{1}\Vert ^{2}+\frac{\sigma \alpha }{2}\Vert u_{2}\Vert ^{2}\leq {c}_{4}e^{-2bz(\theta _{r-\tau }\omega )}. \end{equation} $

对(3.7)式在$ [\tau -t, r] $ ($ r\geq \tau -t $)上应用Gronwall不等式,可得

(3.8) $ \begin{eqnarray} &&\|\psi (r)\|_{E}^{2}+\int_{\tau -t}^{r}e^{\int_{s}^{r}(2bz(\theta _{l-\tau }\omega )-\frac{{\rho }}{2}){\rm d}l}\left( 2\beta\| \nabla u_{1}(s)\|^{2}+\frac{\sigma \alpha }{2}\|u_{2}(s)\|^{2}\right){\rm d}s {} \\ &\leq &e^{\int_{\tau -t}^{r}(2bz(\theta _{l-\tau }\omega )-\frac{{\rho }}{2}){\rm d}l}\|\psi _{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}+{c}_{4}\int_{\tau -t}^{r}e^{-2bz(\theta _{s-\tau }\omega)+\int_{s}^{r}(2bz(\theta _{l-\tau }\omega )-\frac{{\rho }}{2}){\rm d}l}{\rm d}s \end{eqnarray} $

和

$ \begin{eqnarray*} &&\Vert \psi (\tau )\Vert _{E}^{2}+\int_{\tau -t}^{\tau }e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-\frac{{\rho }}{2}){\rm d}l}\left( 2\beta \Vert \nabla u_{1}(s)\Vert ^{2}+\frac{\sigma \alpha }{2}\|u_{2}(s)\|^{2}\right){\rm d}s {\nonumber} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho }}{2}){\rm d}l}\Vert \psi _{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}+{c}_{4}Q_{0}(\omega ). \label{4.4.0} \end{eqnarray*} $

由$ \psi _{\tau -t}(\theta _{-\tau }\omega )\in D(\tau -t, \theta_{-t}\omega ) $,知$ \lim\limits_{t\rightarrow \infty}e^{\int_{-t}^{0}2bz(\theta _{s}\omega ){\rm d}s-\frac{{\rho }}{2}t}\|\psi_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2} = 0 $. (ⅰ)成立.

(ⅱ)根据(3.8)式和$ \psi _{\tau -t}(\theta _{-\tau }\omega )\in D_{0}(\theta _{-t}\omega ) $可得,对任意$ t\geq 1 $有

(3.9) $ \begin{equation} \|\psi (\tau -1)\|_{E}^{2}\leq \, e^{\int_{-t}^{-1}(2bz(\theta _{s}\omega )-\frac{{\rho }}{2}){\rm d}s}L_{0}^{2}(\theta _{-t}\omega )+{c}_{4}\int_{-t}^{-1}e^{-2bz(\theta _{s}\omega ) +\int_{s}^{-1}(2bz(\theta_{l}\omega )-\frac{{\rho }}{2}){\rm d}l}{\rm d}s. \end{equation} $

对(3.7)式在$ [\tau-1, \tau ] $上应用Gronwall不等式,并由(3.9)式,可推出

(3.10) $ \begin{eqnarray} &&\Vert \psi (\tau )\Vert _{E}^{2}+\int_{\tau -1}^{\, \tau }e^{\int_{s}^{\tau}(2bz(\theta _{l-\tau }\omega )-\frac{{\rho }}{2}){\rm d}l}\left( 2\beta \Vert\nabla u_{1}(s)\Vert ^{2}+\frac{\sigma \alpha }{2}\Vert u_{2}(s)\Vert^{2}\right){\rm d}s {} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho }}{2}){\rm d}l}L_{0}^{2}(\theta _{-t}\omega )+{c}_{4}Q_0(\omega). \end{eqnarray} $

由于

$ \begin{eqnarray*} &&\int_{\tau -1}^{\, \tau }e^{\, \int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega)-\frac{{\rho }}{2}){\rm d}l}\left( 2\beta \Vert \nabla u_{1}(s)\Vert ^{2}+\frac{\sigma \alpha }{2}\Vert u_{2}(s)\Vert ^{2}\right){\rm d}s \\ &\geq &e^{-\frac{{\rho }}{2}-\int_{-1}^{0}2b|z(\theta _{l}\omega)|{\rm d}l}\int_{\tau -1}^{\tau }\left( 2\beta\Vert \nabla u_{1}(s)\Vert^{2}+\frac{\sigma\alpha}{2}\Vert u_{2}(s)\Vert ^{2}\right){\rm d}s, \end{eqnarray*} $

从(3.10)式可知

$ \begin{eqnarray*} &&\int_{\tau -1}^{\, \tau }\left( 2\beta \Vert \nabla u_{1}(s)\Vert ^{2}+\frac{\sigma \alpha }{2}\Vert u_{2}(s)\Vert ^{2}\right){\rm d}s \\ &\leq &e^{\frac{{\rho }}{2}+\int_{-1}^{0}2b|z(\theta _{l}\omega)|{\rm d}l+\int_{-t}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho }}{2}){\rm d}l}L_{0}^{2}(\theta _{-t}\omega )+{c}_{4}e^{\, \frac{{\rho }}{2}+\int_{-1}^{0}2b|z(\theta _{l}\omega )|{\rm d}l}Q_{0}(\omega ). \end{eqnarray*} $

进一步有

$ \int_{\tau -1}^{\tau }\left( 2\beta \Vert \nabla u_{1}(s)\Vert ^{2}+\frac{\sigma \alpha }{2}\Vert u_{2}(s)\Vert ^{2}\right){\rm d}s\leq \tilde{Q}_{1}^{2}(\omega ), \qquad \forall t\geq T_{1, D_{0}}(\omega ). $

取内积$ \big((3.2)_1 $, $ -\Delta u_{1}\big) $并由

$ \begin{eqnarray*} && 2\alpha(u_2, \Delta u_1)+2e^{bz(\theta_{r-\tau}\omega)} (g_1, -\Delta u_1)\leq 2\alpha^2\|u_2\|^2+\|\Delta u_1\|^2 +2e^{-2bz(\theta_{r-\tau}\omega)}\|g_1\|^2, \\ &&-2e^{-bz(\theta_{r-\tau}\omega)}\int_{{{\Bbb R}} ^3} f(x, e^{bz(\theta_{r-\tau}\omega)}u_1)\Delta u_1{\rm d}x\\ & = & 2e^{-bz(\theta_{r-\tau}\omega)}\int_{{{\Bbb R}} ^3} \Big(\frac{\partial f}{\partial x}(x, e^{bz(\theta_{r-\tau}\omega)}u_1)\nabla u_1+ e^{bz(\theta_{r-\tau}\omega)}\frac{\partial f}{\partial u}(x, e^{bz(\theta_{r-\tau}\omega)}u_1)|\nabla u_1|^2\Big){\rm d}x\\ &\leq&2e^{-bz(\theta_{r-\tau}\omega)}\|\beta_3\|\, \|\nabla u_1\|+2c_2\|\nabla u_1\|^2\leq e^{-2bz(\theta_{r-\tau}\omega)}\|\beta_3\|^2+(1+2c_2) \|\nabla u_1\|^2, \end{eqnarray*} $

可知,对$ r\geq \tau-t $有

(3.11) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t} \Vert \nabla u_{1}\Vert ^{2}+\Vert \Delta u_{1}\Vert^{2}+\lambda \Vert \nabla u_{1}\Vert^{2} {} \\ &\leq &\left( 1+2c_{2}+2bz(\theta _{r-\tau }\omega )-\lambda \right) \Vert\nabla u_{1}\Vert ^{2}+2\alpha ^{2}\Vert u_{2}\Vert ^{2}+{c}_{5}e^{-2bz(\theta _{r-\tau }\omega )}. \end{eqnarray} $

取$ t\geq {T}_{1, D_{0}}(\omega )\geq 1 $及$ s \in [\tau -1, \tau ] $,在$ [s, \tau] $上对(3.11)式积分可得

$ \begin{eqnarray*} &&\Vert \nabla u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau-t}(\theta _{-\tau }\omega ))\Vert ^{2}-\Vert \nabla u_{1}(s, \tau -t, \theta_{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))\Vert ^{2} \\ &\leq &{c}_{5}\int_{\tau -1}^{\, \tau }e^{-2bz(\theta _{l-\tau }\omega)}{\rm d}l+\int_{\tau -1}^{\, \tau }\left[ \, (1+2c_{2}+2b|z(\theta _{l-\tau }\omega)|)\Vert \nabla u_{1}(l)\Vert ^{2}+2\alpha ^{2}\|u_{2}(l)\|^{2}\, \right]{\rm d}l. \end{eqnarray*} $

对上式在$ [\tau -1, \tau ] $上关于$ s $积分可以推出,对$ t\geq {T}_{1, D_{0}}(\omega ) $有

$ \begin{eqnarray*} &&\Vert \nabla u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))\Vert ^{2} \\ &\leq &{\ c}_{5}\int_{\tau -1}^{\, \tau }e^{-2bz(\theta _{l-\tau }\omega )}{\rm d}l+2\int_{\tau -1}^{\tau }\left[ \, (1+c_{2}+b|z(\theta _{l-\tau }\omega)|)\|\nabla u_{1}(l)\|^{2}+\alpha ^{2}\|u_{2}(l)\|^{2}\, \right]{\rm d}l \\ &\leq &{c}_{5}\int_{-1}^{\, 0}e^{-2bz(\theta _{s}\omega )}{\rm d}s+\left[ \frac{1}{\beta}\left(1+c_{2}+ b\max\limits_{-1\leq s\leq 0}|z(\theta _{s}\omega)|\right) +\frac{4\alpha }{\sigma }\right] \tilde{Q}_{1}^{2}(\omega )\doteq L_{1}^{2}(\omega ), \end{eqnarray*} $

其中$ L_{1}(\omega ) $是只依赖$ |z(\omega)| $的随机变量.

(ⅲ)由(ⅰ)和(ⅱ)可知(ⅲ)成立.引理3.1证毕.

接下来考虑解在$ H^1({{\Bbb R}} ^3)\times L^2( {{\Bbb R}} ^3) $中的尾部估计.

对任意$ \tau \in {{\Bbb R}} $和$ \omega \in \Omega, $令$ T_*(\omega) = T_{D_0}(\omega)+T_{1, D_0}(\omega) $和

(3.12) $ \begin{equation} D_2(\tau - s, \theta_{-s}\omega) = \bigcup\limits_{t\geq \max\{T_{*}(\theta_{-s}\omega), T_{*}(\omega)\}} \Psi(t, \tau - s - t, \theta_{-t-s}\omega)D_0(\theta_{-t-s}\omega), \quad s\geq0, \end{equation} $

从引理3.1知

(3.13) $ \begin{equation} D_{2}(\tau , \omega) \subseteq D_{0}(\omega )\cap D_{1}(\omega)\subset E_{1}\subset E. \end{equation} $

可以断言

(3.14) $ \begin{equation} \Psi(t, \tau-t, \theta_{-t}\omega)D_2(\tau-t, \theta_{-t}\omega) \subseteq D_2(\tau, \omega), \quad \tau\in{{\Bbb R}} , \, \, \omega\in \Omega, \, \, t\geq0. \end{equation} $

事实上,对任意的$ \tau\in{{\Bbb R}} $, $ \omega \in \Omega, $$ t\geq0 $,任取$ v\in D_2(\tau-t, \theta_{-t}\omega) $,那么存在$ \hat{t}\geq\max\{T_{*}(\theta_{-t}\omega), $$ T_{*}(\omega )\} $和$ \hat{v}\in D_0(\theta_{-\hat{t}-t}\omega) $,使得$ v = \Psi(\hat{t}, \tau-t-\hat{t}, \theta_{-\hat{t}-t}\omega)\hat{v}, $因此

$ \begin{eqnarray*} \Psi(t, \tau-t, \theta_{-t}\omega)v& = & \Psi(t, \tau-t, \theta_{-t}\omega) \Psi(\hat{t}, \tau-t-\hat{t}, \theta_{-\hat{t}-t}\omega)\hat{v}\\ & = &\Psi(t+\hat{t}, \tau-t-\hat{t}, \theta_{-\hat{t}-t} \omega)\hat{v}\quad ({\rm let}\ r = t+\hat{t}\, ) \\ &\in& \Psi(r, \tau-r, \theta_{-r}\omega) D_0(\theta_{-r}\omega). \end{eqnarray*} $

由于$ r\geq \max\{T_{*}(\theta_{-t}\omega), T_{*}(\omega) \}+t\geq T_{*}(\omega) $,在(3.12)式中取$ s = 0 $可推出(3.14)式成立.

取函数$ \vartheta \in C^{1}({{\Bbb R}} ^{+}, {{\Bbb R}} ) $使得

$ \begin{eqnarray*} \left\{ \begin{array}{ll} \vartheta (s) = 0, & 0\leq s\leq 1; \\ 0\leq \vartheta (s)\leq 1, & 1\leq s\leq 2; \\ \vartheta (s) = 1, & 2\leq s<+\infty ; \\ |\vartheta ^{\prime }(s)|\leq \tilde{C}, & \forall s\in {{\Bbb R}} ^{+}, \, \, \, \tilde{C}>0, \end{array}\right. \end{eqnarray*} $

其中$ \tilde{C} $为常数.注意,下文中出现的$ c_{i} $ ($ i\in {\Bbb N} $)表示独立于$ (R, \omega , \tau , t) $的正常数.

关于解的尾估计有如下结果.

引理3.2  对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega, $$ R>1 $和$ t\geq 0 $,设$ \psi (r) = (u_{1}(r), u_{2}(r)) = (u_{1}(r, \tau -t, $$ \theta _{-\tau}\omega, \psi_{\tau-t}(\theta _{-\tau}\omega)), u_{2}(r, \tau-t, \theta_{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega))) $$ (r\geq \tau -t) $是系统(3.2)的具有初值$ \psi _{\tau-t}(\theta _{-\tau }\omega )\in D_{2}(\tau -t, \theta _{-t}\omega) $的解,那么

(ⅰ)

(3.15) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\left( \beta |u_{1}(\tau )|^{2}+\alpha |u_{2}(\tau )|^{2}\right) {\rm d}x {} \\ &&+2\beta \int_{\tau -t}^{\tau }e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}(s)|^{2}{\rm d}x{\rm d}s {} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{s}\omega )-{\rho }){\rm d}s}L_{2}^{2}(\theta _{-t}\omega )+\frac{Q_{1}(\omega )}{R}+\zeta _{1, R}Q_{0}(\omega ), \end{eqnarray} $

其中

$ Q_{1}(\omega ) = 2\sqrt{2\beta }\tilde{C}\int_{-\infty }^{0}e^{\int_{s}^{0}(2bz(\theta _{l}\omega )-{\rho }){\rm d}l}L_{2}^{2}(\theta_{s}\omega){\rm d}s, $

$ \zeta _{1, R} = 2\beta \int_{|x|\geq R}\beta _{1}(x){\rm d}x+\frac{\beta }{\lambda }\sup\limits_{s\in {{\Bbb R}} }\int_{|x|\geq R}g_{1}^{2}(x, s){\rm d}x+\frac{2\alpha }{3\sigma }\sup\limits_{s\in {{\Bbb R}} }\int_{|x|\geq R}g_{2}^{2}(x, s){\rm d}x; $

(ⅱ)存在正常数$ {c}_{7} $和随机变量$ Q_{2}(\omega )>0 $,使得

(3.16) $ \begin{equation} \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}(\tau )|^{2}{\rm d}x\leq {c}_{7}e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-{\rho}){\rm d}l}(1+t)L_{2}^{2}(\theta _{-t}\omega )+\frac{Q_{2}(\omega )}{R}+\zeta_{2, R}Q_{0}(\omega ), \end{equation} $

其中

$ \zeta_{2, R} = \zeta_{3, R}+(\frac{1}{2\beta}+\frac{2\alpha }{{\rho }})\zeta_{1, R}, \quad \zeta _{3, R} = \sup\limits_{s\in {{\Bbb R}} }\int_{|x|\geq R}g_{1}^{2}(x, s){\rm d}x+\int_{|x|\geq R}\beta _{3}^{2}(x){\rm d}x; $

(ⅲ)对任意$ \varepsilon >0 $,存在$ {T}_{\varepsilon }(\omega )>0 $,使得

$ \begin{eqnarray*} &&\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\left( \beta |u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau}\omega ))|^{2}+\alpha |u_{2}(\tau , \tau -t, \theta _{-\tau }\omega , \psi_{\tau -t}(\theta _{-\tau }\omega ))|^{2}\right. \\ &&+\left. \, |\nabla u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi_{\tau -t}(\theta _{-\tau }\omega ))|^{2}\right) {\rm d}x \\&\leq &\varepsilon +\frac{Q_{1}(\omega )+Q_{2}(\omega )}{R}+\left(\zeta_{3, R}+(\frac{1}{2\beta }+\frac{2\alpha }{\rho }+1)\zeta _{1, R}\right) Q_{0}(\omega ), \qquad \forall \, t\geq {T}_{\varepsilon }(\omega ). \end{eqnarray*} $

证  从(3.13)和(3.14)式知,对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega $, $ t\geq 0 $,

(3.17) $ \begin{equation} \left\{\begin{array}{l} \psi (r)\in D_{2}(r, \theta _{r-\tau }\omega )\subseteq D_{0}(\theta _{r-\tau}\omega )\cap D_{1}(\theta _{r-\tau }\omega )\subset E_{1}, \\ \|\psi (r)\|_{E_{1}}^{2} = \beta \Vert u_{1}(r)\Vert ^{2}+\alpha \Vert u_{2}(r)\Vert ^{2}+\Vert \nabla u_{1}(r)\Vert ^{2}\leq L_{2}^{2}(\theta _{r-\tau }\omega ), \end{array} \right. r\geq \tau -t. \end{equation} $

(ⅰ)取内积$ \big( $$ (3.2)_1 $, $ \beta \vartheta (\frac{|x|^{2}}{R^{2}})u_{1} $$ \big) $和$ \big( $$ (3.2)_2 $, $ \alpha \vartheta (\frac{|x|^{2}}{R^{2}})u_{2} $$ \big) $并相加,再由以下估计

$ 2\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})u_{1}\Delta u_{1}{\rm d}x\leq -2\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}} )|\nabla u_{1}|^{2}{\rm d}x+\frac{2\sqrt{2\beta }\tilde{C}}{R}L_{2}^{2}(\theta _{r-\tau }\omega ), $

$ 2\beta e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1})\vartheta (\frac{|x|^{2}}{R^{2}})u_{1}{\rm d}x\leq 2\beta e^{-2bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\beta _{1}(x){\rm d}x, $

$ 2\beta e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}g_{1}\vartheta (\frac{|x|^{2}}{R^{2}})u_{1}{\rm d}x\leq \lambda \beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|u_{1}|^{2}{\rm d}x+\frac{\beta }{\lambda }e^{-2bz(\theta _{r-\tau }\omega)}\int_{{{\Bbb R}} ^{3}} \vartheta (\frac{|x|^{2}}{R^{2}})g_{1}^{2}{\rm d}x, $

$ 2\alpha e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}g_{2}\vartheta (\frac{|x|^{2}}{R^{2}})u_{2}{\rm d}x\leq \frac{3\sigma \alpha }{2}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|u_{2}|^{2}{\rm d}x+\frac{2\alpha }{3\sigma}e^{-2bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})g_{2}^{2}{\rm d}x, $

可得

(3.18) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\left( \beta |u_{1}|^{2}+\alpha |u_{2}|^{2}\right) {\rm d}x+({\rho }-2bz(\theta _{r-\tau}\omega))\int_{{{\Bbb R}} ^{3}} \vartheta (\frac{|x|^{2}}{R^{2}})(\beta |u_{1}|^{2}+\alpha |u_{2}|^{2}){\rm d}x{} \\ &\leq &-2\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x+\frac{2\sqrt{2\beta }\tilde{C}}{R}L_{2}^{2}(\theta _{r-\tau }\omega ) {} \\ &&+e^{-2bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\left( 2\beta \beta _{1}(x)+\frac{\beta }{\lambda}g_{1}^{2}(x, r)+\frac{2\alpha }{3\sigma }g_{2}^{2}(x, r)\right){\rm d}x. \end{eqnarray} $

对(3.18)式在$ [\tau -t, \tau ] $上应用Gronwall不等有

(3.19) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\left( \beta |u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau}\omega ))|^{2}+\alpha |u_{2}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))|^{2}\right){\rm d}x{} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{s}\omega )-{\rho }){\rm d}s}\Vert \psi _{\tau-t}(\theta _{-\tau }\omega )\Vert _{E}^{2}{} \\ && -2\beta \int_{\tau -t}^{\, \tau }e^{\int_{s}^{\tau }(2bz(\theta_{l-\tau }\omega )-{\rho }){\rm d}l}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{ R^{2}})|\nabla u_{1}(s, \tau -t, \theta _{-\tau }\omega , \psi _{\tau-t}(\theta _{-\tau }\omega ))|^{2}{\rm d}x{\rm d}s{} \\ &&+\frac{2\sqrt{2\beta }\tilde{C}}{R}\int_{\tau -t}^{\tau}e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-{\rho}){\rm d}l}L_{2}^{2}(\theta _{s-\tau }\omega ){\rm d}s+\, \int_{\tau -t}^{\tau}e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l-2bz(\theta _{s-\tau }\omega)}{} \\ & &\times \left( 2\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{ |x|^{2}}{R^{2}})\beta _{1}(x){\rm d}x+\frac{\beta }{\lambda }\mathop{\sup} \limits_{s\in {{\Bbb R}} }\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})g_{1}^{2}{\rm d}x+\frac{2\alpha }{3\sigma }\mathop{\sup}\limits_{s\in {{\Bbb R}} }\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})g_{2}^{2}{\rm d}x\right){\rm d}s{} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{s}\omega )-{\rho }){\rm d}s}L_{2}^{2}(\theta_{-t}\omega ) -2\beta \int_{\tau -t}^{\, \tau }e^{\int_{s}^{\tau }(2bz(\theta_{l-\tau }\omega )-{\rho }){\rm d}l}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}(s)|^{2}{\rm d}x{\rm d}s {} \\ &&+\frac{2\sqrt{2\beta }\tilde{C}}{R}\int_{\tau -t}^{\tau}e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-{\rho}){\rm d}l}L_{2}^{2}(\theta _{s-\tau }\omega ){\rm d}s+\, \int_{\tau -t}^{\tau}e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-\frac{{\rho }}{2} ){\rm d}l-2bz(\theta _{s-\tau }\omega )}{} \\ &&\times \left( 2\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{ |x|^{2}}{R^{2}})\beta _{1}(x){\rm d}x+\frac{\beta }{\lambda }\mathop{\sup}\limits_{s\in{{\Bbb R}} }\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})g_{1}^{2}{\rm d}x+\frac{2\alpha }{3\sigma }\mathop{\sup}\limits_{s\in {{\Bbb R}} }\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})g_{2}^{2}{\rm d}x\right){\rm d}s{} \\ &\leq & e^{\int_{-t}^{0}(2bz(\theta _{s}\omega )-{ \rho }){\rm d}s}L_{2}^{2}(\theta _{-t}\omega )-\, 2\beta \int_{\tau -t}^{\, \tau }e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-{\rho}){\rm d}l}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}(s)|^{2}{\rm d}x{\rm d}s {} \\ && +\frac{Q_{1}(\omega )}{R}+\zeta _{1, R}Q_{0}(\omega). \end{eqnarray} $

(ⅱ)取内积$ \big( $$ (3.2)_1 $, $ -\vartheta (\frac{|x|^{2}}{R^{2}})\Delta u_{1} $$ \big) $得

(3.20) $ \begin{eqnarray} &&-\int_{{{\Bbb R}} ^{3}}\frac{{\rm d}u_{1}}{{\rm d}t} \vartheta (\frac{|x|^{2}}{R^{2}})\Delta u_{1}{\rm d}x+\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\Delta u_{1}|^{2}{\rm d}x {} \\ &&-(\lambda -bz(\theta _{r-\tau }\omega ))\int_{{{\Bbb R}} ^{3}}\vartheta(\frac{|x|^{2}}{R^{2}})u_{1}\Delta u_{1}{\rm d}x-\alpha \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})u_{2}\Delta u_{1}{\rm d}x {} \\ & = &-e^{-bz(\theta _{r-\tau }\omega )}\left( \int_{{{\Bbb R}} ^{3}}\vartheta( \frac{|x|^{2}}{R^{2}})f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1})\Delta u_{1}{\rm d}x+\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\Delta u_{1}g_{1}{\rm d}x\right) . \end{eqnarray} $

令$ \bar{\beta} = \min \{1, \sqrt{\beta }\}>0 $.根据(3.1)和(3.17)式,可得

(3.21) $ \begin{eqnarray} \|u_{1}(r, \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau}\omega ))\|_{L^{6}({{\Bbb R}} ^{3})} &\leq& C_{0}\left( \Vert \nabla u_{1}(r)\Vert ^{2}+\Vert u_{1}(r)\Vert ^{2}\right) ^{\frac{1}{2}}{}\\ &\leq& C_{0} \bar{\beta}^{-1}L_{2}(\theta _{r-\tau }\omega ). \end{eqnarray} $

考虑(1.2)和(3.21)式,有

(3.22) $ \begin{eqnarray} &&\|e^{-bz(\theta _{r-\tau }\omega )}f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}(r))\|^{2} {} \\ &\leq &e^{-2bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\left( c_{1}|e^{bz(\theta _{r-\tau }\omega )}u_{1}(r)|^{3}+\beta _{2}(x)\right)^{2}{\rm d}x{} \\ &\leq &2c_{1}^{2}C_{0}^{6}\bar{\beta}^{-6}e^{4bz(\theta _{r-\tau }\omega)}L_{2}^{6}(\theta _{r-\tau }\omega )+2\|\beta _{2}\|^{2}e^{-2bz(\theta_{r-\tau }\omega )}. \end{eqnarray} $

由$ (3.2)_1 $式知

$ \begin{eqnarray*} \left\|\frac{{\rm d}u_{1}}{{\rm d}t}\right\|^{2}&\leq &5\|\Delta u_{1}\|^{2}+{c}_{8}(1+z^{2}(\theta_{r-\tau }\omega ))L_{2}^{2}(\theta_{r-\tau }\omega)+ {c}_{9}e^{4bz(\theta_{r-\tau}\omega)}L_{2}^{6}(\theta _{r-\tau}\omega)\\ &&+5e^{-2bz(\theta_{r-\tau }\omega )}\left(2\|\beta_{2}\|^{2}+\|g_{1}\|^{2}\right) . \end{eqnarray*} $

因此

(3.23) $ \begin{eqnarray} &&-\int_{{{\Bbb R}} ^{3}}\frac{{\rm d}u_{1}(t)}{{\rm d}t}\vartheta (\frac{|x|^{2}}{R^{2}})\Delta u_{1}{\rm d}x = \int_{{{\Bbb R}} ^{3}}\nabla u_1\cdot\left( \frac{2x}{R^2}\vartheta'(\frac{|x|^{2}}{R^{2}})\frac{{\rm d}u_1}{{\rm d}t} +\vartheta(\frac{|x|^{2}}{R^{2}})\nabla(\frac{{\rm d}u_1}{{\rm d}t}) \right){\rm d}x{} \\&\geq& -\frac{\sqrt{2}\tilde{C}}{R}\left(\|\nabla u_1\|^2 +\|\frac{{\rm d}u_1}{{\rm d}t}\|^2\right)+\frac{1}{2}\frac{\rm d}{{\rm d}t} \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x{} \\&\geq&\frac{1}{2}\frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x-\frac{{c}_{10}}{R}\|\Delta u_{1}\|^{2}-\frac{{c}_{12}}{R}e^{-2bz(\theta _{r-\tau }\omega)}{} \\&&-\frac{{c}_{11}}{R}\left( L_{2}^{2}(\theta_{r-\tau }\omega )+z^{2}(\theta _{r-\tau }\omega )L_{2}^{2}(\theta _{r-\tau }\omega )+e^{4bz(\theta _{r-\tau}\omega )}L_{2}^{6}(\theta _{r-\tau }\omega )\right). \end{eqnarray} $

类似(3.19)式,对(3.18)式在$ [\tau -t, r] $上应用Gronwall不等式得到

(3.24) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\alpha |u_{2}(r, \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))|^{2}{\rm d}x {} \\ &\leq &\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})(\beta |u_{1}(r)|^{2}+\alpha |u_{2}(r, \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))|^{2}){\rm d}x {} \\ &\leq &e^{\int_{\tau -t}^{r}(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l}L_{2}^{2}(\theta _{-t}\omega )+\frac{2\sqrt{2\beta }\tilde{C}}{R}\int_{\tau -t}^{\, r}e^{\int_{s}^{r}(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l}L_{2}^{2}(\theta _{s-\tau }\omega ){\rm d}s{} \\ &&+\int_{\tau -t}^{\, r}e^{\int_{s}^{r}(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l-2bz(\theta _{s-\tau }\omega )}\zeta _{1, R}{\rm d}s. \end{eqnarray} $

因为$ 2ab\leq a^2+b^2 $,从而

(3.25) $ \begin{equation} \alpha \int_{{{\Bbb R}} ^{3}}u_{2}\vartheta (\frac{|x|^{2}}{R^{2}})\Delta u_{1}{\rm d}x\leq \frac{\alpha ^{2}}{2}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})u_{2}^{2}{\rm d}x+\frac{1}{2}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\Delta u_{1}|^{2}{\rm d}x. \end{equation} $

注意到

(3.26) $ \begin{eqnarray} &&(\lambda - bz(\theta _{r-\tau }\omega ))\int_{{{\Bbb R}} ^{3}}\vartheta ( \frac{|x|^{2}}{R^{2}})u_{1}\Delta u_{1}{\rm d}x {} \\ &\leq &(bz(\theta _{r - \tau }\omega ) - \lambda )\int_{{{\Bbb R}} ^{3}} \vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x + (b|z(\theta _{r - \tau }\omega )| + \lambda )\frac{\sqrt{2}\tilde{C}(1 + \beta ^{-1})}{R}L_{2}^{2}(\theta _{r - \tau }\omega ), {\qquad} \end{eqnarray} $

(3.27) $ \begin{eqnarray} &&-e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1})\vartheta (\frac{|x|^{2}}{R^{2}})\Delta u_{1}{\rm d}x {} \\ & = &e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\nabla u_{1}\cdot \frac{\partial f}{\partial x}(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}){\rm d}x {} \\ &&+\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\frac{\partial f}{\partial u}(x, e^{bz(\theta _{r-\tau }\omega )}u_{1})|\nabla u_{1}|^{2}{\rm d}x {} \\ &&+e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\nabla u_{1}\cdot f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1})\vartheta ^{\prime }(\frac{|x|^{2}}{R^{2}})\frac{2x}{R^{2}}{\rm d}x {} \\ &\leq &\frac{1}{2}\, e^{-2bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\beta _{3}(x)|^{2}{\rm d}x+\frac{1}{2}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x {} \\ &&+c_{2}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x+\frac{\sqrt{2}\tilde{C}}{R}\left( \Vert \nabla u_{1}\Vert ^{2}+\Vert e^{-bz(\theta _{r-\tau }\omega )}f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1})\Vert ^{2}\right) {} \\ &\leq &\frac{1}{2}\, e^{-2bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\beta _{3}^{2}(x){\rm d}x+{c}_{13}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x {} \\ &&+\frac{{c}_{14}}{R}\left( L_{2}^{2}(\theta _{r-\tau }\omega )+e^{4bz(\theta _{r-\tau }\omega )}L_{2}^{6}(\theta _{r-\tau }\omega )+e^{-2bz(\theta _{r-\tau }\omega )}\right) , \end{eqnarray} $

(3.28) $ \begin{eqnarray} &&-e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})g_{1}(x, r)\Delta u_{1}{\rm d}x {} \\ &\leq &\frac{1}{2}e^{-2bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})g_{1}^{2}(x, r){\rm d}x+\frac{1}{2}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\Delta u_{1}|^{2}{\rm d}x. \end{eqnarray} $

考虑(3.23)–(3.28)式,再由(3.20)式可以推出

(3.29) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x {}\\ &\leq &2(bz(\theta _{r-\tau }\omega )-\lambda )\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x+\frac{{c}_{15}}{R}\|\Delta u_{1}\|^{2} {} \\ && +\frac{{c}_{16}}{R}\left( L_{2}^{2}(\theta _{r-\tau }\omega )+z^{2}(\theta _{r-\tau }\omega )L_{2}^{2}(\theta _{r-\tau }\omega )+e^{4bz(\theta _{r-\tau }\omega )}L_{2}^{6}(\theta _{r-\tau }\omega )\right) {} \\ && +\left( \zeta _{3, R}+\frac{{c}_{17}}{R}\right) e^{-2bz(\theta _{r-\tau }\omega )}+\alpha e^{\int_{\tau -t}^{r}(2bz(\theta _{s-\tau }\omega )-{\rho }){\rm d}s}L_{2}^{2}(\theta _{-t}\omega ) {} \\ && +\frac{{c}_{18}}{R}\int_{\tau -t}^{r}e^{\int_{s}^{r}(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l}L_{2}^{2}(\theta _{s-\tau }\omega ){\rm d}s+{c}_{19}\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}|^{2}{\rm d}x {} \\ && +\alpha \zeta _{1, R}\int_{\tau -t}^{r}e^{\int_{s}^{r}(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l-2bz(\theta _{s-\tau }\omega )}{\rm d}s. \end{eqnarray} $

对(3.11)式在$ [\tau -t, \tau ] $上应用Gronwall不等式有

(3.30) $ \begin{eqnarray} &&\|\nabla u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))\|^{2}+\int_{\tau -t}^{\tau }e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-\lambda ){\rm d}l}\|\Delta u_{1}(s)\|^{2}{\rm d}s {} \\ & \leq &e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-\lambda ){\rm d}l}L_{2}^{2}(\theta _{-t}\omega ) + (1 + 2c_{2} + 2\alpha )\int_{-\infty }^{0} e^{\int_{s}^{0}(2bz(\theta _{l}\omega ) - \lambda ){\rm d}l}L_{2}^{2}(\theta _{s}\omega ){\rm d}s + {c}_{5}Q_{0}(\omega ).{}\\ \end{eqnarray} $

注意到

(3.31) $ \begin{eqnarray} &&\int_{\tau -t}^{\tau }e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-2\lambda ){\rm d}l+\int_{\tau -t}^{s}(2bz(\theta _{r-\tau }\omega )-{\rho }){\rm d}r}L_{2}^{2}(\theta _{-t}\omega ){\rm d}s {} \\ &\leq &\int_{\tau -t}^{\tau }e^{\int_{\tau -t}^{\tau }(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l}L_{2}^{2}(\theta _{-t}\omega ){\rm d}s = e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-{\rho }){\rm d}l}tL_{2}^{2}(\theta _{-t}\omega ), \end{eqnarray} $

(3.32) $ \begin{eqnarray} &&\int_{\tau -t}^{\tau }e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-2\lambda ){\rm d}l}\int_{\tau -t}^{s}e^{\int_{r}^{s}(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l}L_{2}^{2}(\theta _{r-\tau }\omega ){\rm d}r{\rm d}s {} \\ &\leq &\frac{2}{{\rho }}\int_{-t}^{0}e^{\int_{r}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho }}{2}){\rm d}l}L_{2}^{2}(\theta _{r}\omega ){\rm d}r<\infty , \end{eqnarray} $

(3.33) $ \begin{eqnarray} &&\int_{\tau -t}^{\tau }e^{\int_{s}^{\tau }(2bz(\theta _{l-\tau }\omega )-2\lambda ){\rm d}l}\int_{\tau -t}^{s}e^{\int_{r}^{s}(2bz(\theta _{l-\tau }\omega )-{\rho }){\rm d}l-2bz(\theta _{r-\tau }\omega )}{\rm d}r{\rm d}s {} \\ &\leq &\frac{2}{{\rho }}\int_{-t}^{0}e^{\int_{s}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho }}{2}){\rm d}l-2bz(\theta _{s}\omega )}{\rm d}s<\infty . \end{eqnarray} $

对(3.29)式在$ [\tau -t, \tau ] $上应用Gronwall不等式并由(3.15)式, (3.30)–(3.33)式,可得

$ \begin{eqnarray*} &&\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|\nabla u_{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{\tau -t}(\theta _{-\tau }\omega ))|^{2}{\rm d}x \\ &\leq &{c}_{20}e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-{\rho} ){\rm d}l}(1+t)L_{2}^{2}(\theta _{-t}\omega ) \\ &&+\frac{{c}_{21}}{R}\int_{-\infty }^{0}e^{\int_{s}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho }}{2}){\rm d}l}\left( L_{2}^{2}(\theta _{s}\omega )+z^{2}(\theta _{s}\omega )L_{2}^{2}(\theta _{s}\omega )+e^{4bz(\theta _{s}\omega )}L_{2}^{6}(\theta _{s}\omega )\right){\rm d}s \\ &&+\frac{{c}_{22}}{R}\left( Q_{0}(\omega )+Q_{1}(\omega )\right) +\left( \zeta _{3, R}+(\frac{1}{2\beta }+\frac{2\alpha }{{\rho }})\zeta _{1, R}\right) Q_{0}(\omega ) \\ &\leq &{c}_{7}e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-{\rho }){\rm d}l}(1+t)L_{2}^{2}(\theta _{-t}\omega )+\frac{{c}_{23}}{R}\left( Q_{0}(\omega )+Q_{1}(\omega )+Q_{3}(\omega )\right) +\zeta _{2, R}Q_{0}(\omega ), \end{eqnarray*} $

其中

$ Q_{3}(\omega ) = \int_{-\infty }^{0}e^{\int_{s}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho }}{2}){\rm d}l}\left( L_{2}^{2}(\theta _{s}\omega )+z^{2}(\theta _{s}\omega )L_{2}^{2}(\theta _{s}\omega )+e^{4bz(\theta _{s}\omega )}L_{2}^{6}(\theta _{s}\omega )\right){\rm d}s, $

$ \zeta _{2, R} = \zeta _{3, R}+(\frac{1}{2\beta }+\frac{2\alpha }{\rho} )\zeta _{1, R}. $

令

(3.34) $ \begin{equation} Q_{2}(\omega ) = {c}_{23}\left( Q_{0}(\omega )+Q_{1}(\omega)+Q_{3}(\omega )\right), \end{equation} $

那么(3.16)式成立.

(ⅲ)由于$ \lim\limits_{t\rightarrow\infty}(1+{c}_{7})e^{\int_{-t}^{0}(2bz(\theta _{l}\omega )-{\rho } ){\rm d}l}(1+t)L_{2}^{2}(\theta _{-t}\omega ) = 0 $,取

$ {T}_{\varepsilon }(\omega ) = \min \left\{ t:(1+{c}_{7})e^{\int_{-t}^{0}(2bz( \theta _{l}\omega )-{\rho }){\rm d}l}(1+t)L_{2}^{2}(\theta _{-t}\omega )\leq \varepsilon \right\} <\infty , $

由(ⅰ)和(ⅱ)可知(ⅲ)成立.引理3.2证毕.

接下来考虑随机指数吸引子的存在性.

对任意$ \omega \in \Omega $, $ s\geq0 $, $ \varepsilon>0 $,令

$ {\chi}(\tau-s , \theta_{-s}\omega ) = \overline{\bigcup\limits_{t\geq\max\{T_{\ast }(\omega), T_{\ast }(\theta_{-s}\omega), T_{\ast }(\theta_{-T_{\varepsilon}(\omega)}\omega)\}+ T_{\varepsilon}(\omega) }\Psi (t, \tau-s-t, \theta _{-t-s}\omega )D_{0}(\theta _{-t-s}\omega)}, $

则$ \{{\chi }(\tau , \omega )\}_{\tau \in {{\Bbb R}} , \, \omega \in \Omega } $是$ E = L^{2}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3}) $的缓增随机闭子集族,具有以下性质:

(A)对任意$ \tau \in {{\Bbb R}} , $$ \omega \in \Omega , $$ {\chi} (\tau , \omega )\subseteq D_{0}(\omega )\cap D_{1}(\omega )\subset E^{1} \subset E, $且$ {\chi} (\tau , \omega ) $在$ E $中的直径小于等于$ 2L_{0}(\omega) $,且$ L_{0}(\theta _{t}\omega ) $关于$ t\in{{\Bbb R}} $连续；

(B) $ {\chi }(\tau , \omega ) $是正不变的,即$ \Psi (t, \tau -t, \theta _{-t}\omega ){\chi }(\tau -t, \theta_{-t}\omega )\subseteq{\chi }(\tau , \omega ) $, $ \forall\omega\in\Omega $, $ t\geq0 $.证明类似(3.14)式;

(C)对任意$ \tau \in {{\Bbb R}} , $$ \omega \in \Omega, $$ R>1 $和$ \psi = (u_{1}, u_{2})\in \chi(\tau , \omega ) $,下面的不等式成立

$ \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\left( |\nabla u_{1}|^{2}+\beta |u_{1}|^{2}+\alpha |u_{2}|^{2}\right) {\rm d}x\leq \varepsilon + \frac{Q_{1}(\omega )+Q_{2}(\omega )}{R}+\zeta _{4, R}Q_{0}(\omega ), \label{4.17.0} $

其中$ \zeta _{4, R} = \zeta _{3, R}+(\frac{1}{2\beta }+\frac{2\alpha }{\bar{\rho}}+1)\zeta _{1, R}. $事实上,对任意$ v\in\chi(\tau, \omega) $,存在$ \hat{v}\in D_0(\theta_{-\hat{t}}\omega) $和$ \hat{t}\geq\max\{T_{\ast }(\omega), T_{\ast}(\theta_ {-T_{\varepsilon}(\omega)}\omega)\}+T_{\varepsilon}(\omega) $使得

$ \begin{eqnarray*} v& = &\Psi(\hat{t}, \tau-\hat{t}, \theta_{-\hat{t}}\omega)\hat{v} \\&\in&\Psi(T_{\varepsilon}(\omega), \tau-T_{\varepsilon}(\omega), \theta_{-T_{\varepsilon}(\omega)}\omega) \Psi(\hat{t}-T_{\varepsilon}(\omega), \tau-\hat{t}, \theta_{-\hat{t}}\omega)D_0(\theta_{-\hat{t}}\omega)\\ & \subseteq&\Psi(T_{\varepsilon}(\omega), \tau- T_{\varepsilon}(\omega), \theta_{-T_{\varepsilon}(\omega)}\omega) D_2(\tau-T_{\varepsilon}(\omega), \theta_{-T_{\varepsilon}(\omega)}\omega), \end{eqnarray*} $

由引理3.2 (ⅲ)知(C)成立.

为证明$ \Psi $存在随机指数吸引子,需验证$ \{{\chi }(\tau , \omega )\}_{\tau \in {{\Bbb R}} , \, \omega \in \Omega } $满足定理2.1的(H2)–(H3).

接下来验证$ \Psi $的Lipschitz性.

对任意的$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega $, $ t\geq 0 $和$ \psi_{j, \tau -t}(\theta _{-\tau }\omega ) = (u_{1, \tau -t}^{(j)}(\theta_{-\tau }\omega), u_{2, \tau -t}^{(j)}(\theta _{-\tau }\omega ))\in{\chi }(\tau -t, $$ \theta _{-t}\omega ) $, $ j = 1, $$ 2, $令

$ \psi_{j}(r) = \psi_{j}(r, \tau -t, \theta _{-\tau }\omega , \psi_{j, \tau -t}(\theta _{-\tau }\omega )) = (u_{1 }^{(j)}(r ), \, u_{2 }^{(j)}(r)), $

$ y(r) = \psi_{1}(r)-\psi_{2}(r) = (y_{1 } (r ), \, y_{2 } (r)), \quad r\geq \tau -t, \quad j = 1, 2. \label{3.57} $

考虑到$ {\chi }(\tau , \omega ) $的正不变性和性质(A),可以推出

(3.35) $ \begin{equation} \left\{ \begin{array}{l} \psi _{j}(r) = (u_{1}^{(j)}(r), \, u_{2}^{(j)}(r))\in {\chi }(r, \theta _{r-\tau }\omega )\subseteq D_{1}(\theta _{r-\tau }\omega )\subset E_{1}, \\ \|\psi_{j}(r)\|_{E_{1}}^{2} = \|\nabla u_{1}^{(j)}(r)\|^{2}+\beta \|u_{1}^{(j)}(r)\|^{2}+\alpha\|u_{2}^{(j)}(r)\|^{2}\leq L_{2}^{2}(\theta_{r-\tau }\omega ), \end{array} \right. \ \forall r\geq \tau -t, j = 1, 2. \end{equation} $

基于(3.2)式,有

(3.36) $ \begin{equation} \left\{ \begin{array}{l} { } \frac{{\rm d}y_{1}}{{\rm d}t} = \Delta y_1 - \lambda y_1+bz(\theta _{r-\tau }\omega )y_1 - \alpha y_2\\ {\qquad}{\quad}+e^{-bz(\theta _{r-\tau }\omega )}[f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(1)}) - f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(2)})], \\ { } \frac{{\rm d}y_{2}}{{\rm d}t} = -\sigma y_2 + bz(\theta _{r-\tau }\omega )y_2 + \beta y_1, \quad r\geq \tau -t, \tau \in {{\Bbb R}} , \\ y_j(\tau - t, \tau - t, \theta _{-\tau }\omega , y_{j, \tau -t}(\theta _{-\tau }\omega )) = u^{(1)}_{j, \tau - t}(\theta _{-\tau }\omega ) - u^{(2)}_{j, \tau - t}(\theta _{-\tau }\omega ), \ j = 1, 2. \end{array} \right. \end{equation} $

下面的引理说明$ \Psi $在$ {\chi }(\tau , \omega ) $上具有Lipschitz性.

引理3.3  对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega, $$ t\geq 0 $和$ \psi_{j, \tau -t}(\theta _{-\tau }\omega)\in {\chi}(\tau -t, \theta _{-t}\omega ) $, $ j = 1, 2, $有下面的不等式成立

(3.37) $ \begin{eqnarray} &&\|\psi _{1}(r, \tau -t, \theta _{-\tau }\omega , \psi _{1, \tau -t}(\theta _{-\tau }\omega ))-\psi _{2}(r, \tau -t, \theta _{-\tau }\omega , \psi _{2, \tau -t}(\theta _{-\tau }\omega ))\|_{E}^{2} {} \\ && +2\beta \int_{\tau -t}^{\, r}e^{\int_{s}^{r}2(bz(\theta _{l-\tau }\omega )+c_{2}){\rm d}l}\|\nabla y_{1}(s)\|^{2}{\rm d}s {} \\ &\leq &e^{\int_{\tau -t}^{r}2(bz(\theta _{l-\tau }\omega )+c_{2}){\rm d}l}\, \| \psi _{1, \tau -t}(\theta _{-\tau }\omega )-\psi _{2, \tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}, \quad \forall r\geq \tau -t. \end{eqnarray} $

特别地,有

(3.38) $ \begin{eqnarray} &&\Vert \psi _{1}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{1, \tau -t}(\theta _{-\tau }\omega ))-\psi _{2}(\tau , \tau -t, \theta _{-\tau }\omega , \psi _{2, \tau -t}(\theta _{-\tau }\omega ))\Vert _{E} {} \\ &\leq &e^{\int_{-t}^{0}(bz(\theta _{l}\omega )+c_{2}){\rm d}l}\, \Vert \psi _{1, \tau -t}(\theta _{-\tau }\omega )-\psi _{2, \tau -t}(\theta _{-\tau }\omega )\Vert _{E}, \quad \forall r\geq \tau -t. \end{eqnarray} $

证  取内积$ \big((3.36)_1, $$ \beta y_1(r)\big) $和$ \big((3.36)_2, $$ \alpha y_2(r)\big) $并相加,再由

$ 2\beta \, e^{-b z(\theta _{r-\tau}\omega )}\int_{{{\Bbb R}} ^{3}}[f(x, e^{b z(\theta _{r-\tau}\omega )}u_{1}^{(1)}(r))-f(x, e^{b z(\theta _{r-\tau}\omega )}u_{1}^{(2)}(r))]y_1(r){\rm d}x\leq2\beta c_{2}\|y_1(r)\|^{2}, $

可以得到

$ \frac{\rm d}{{\rm d}t}\Vert y(r)\Vert _{E}^{2}+2 \beta \Vert \nabla y_1(r)\Vert ^{2} \leq 2(b z(\theta _{r-\tau}\omega )+c_{2})\|y(r)\|_{E}^{2}, \quad \forall r\geq \tau -t. $

对上式应用Gronwall不等式推出(3.37)–(3.38)式成立.引理3.3证毕.

接下来对解进行分解.考虑特征值问题

$ \left\{ \begin{array}{l} -\Delta \tilde{u}(x) = \mu \tilde{u}(x), \qquad x\in {\Bbb U}_{2R} , \\ \tilde{u}(x) = 0, \qquad \qquad \quad \, \, \, \, x\in\partial \, {\Bbb U} _{2R}, \end{array} \right. $

其中$ {\Bbb U}_{R} = \{x\in {{\Bbb R}} ^{3}:|x|<R\} $表示以原点为球心以$ 0<R<+\infty $为半径的球.由文献[28]知,存在一簇特征函数$ \{\tilde{e}_{m, R}\}_{m\in {\Bbb N}} $和对应的一簇特征值$ \{\mu _{m, R}\}_{m\in {\Bbb N}} $,使得这簇特征函数构成$ L^{2}({\Bbb U}_{2R}) $和$ H_{0}^{1}({\Bbb U}_{2R}) $的标准正交基,这簇特征值$ \{\mu _{m, R}\}_{m\in {\Bbb N}} $满足

$ 0<\mu _{1, R}\leq \mu _{2, R}\leq \cdots \leq \mu _{m, R}\leq \cdots , \quad \mu _{m, R}\rightarrow +\infty \quad{\rm as}\quad m\rightarrow +\infty \, . \label{S} $

此外,对于给定的$ R\in (0, +\infty ) $和任意$ m\in {\Bbb N} $,

$ -\Delta \tilde{e}_{m, R} = \mu _{m, R}\tilde{e}_{m, R}, \quad \tilde{e}_{m, R}\in H^{2}({\Bbb U}_{2R})\cap H_{0}^{1}({\Bbb U}_{2R}). $

对于给定的$ m\in {\Bbb N} $, $ L^{2}({\Bbb U}_{2R}) = L_{m}^{2}({\Bbb U}_{2R})+L_{m}^{2}({\Bbb U}_{2R})^{\bot } $,其中

$ L_{m}^{2}({\Bbb U}_{2R}) = {\rm span}\{\tilde{e}_{1, R}, \tilde{e}_{2, R}, \cdots , \tilde{e}_{m, R}\}, \quad L_{m}^{2}({\Bbb U}_{2R})^{\bot} = {\rm span}\{\tilde{e}_{m+1, R}, \tilde{e}_{m+2, R}, \cdots \}. $

令

$ \tilde{P}_{m, R}:L^{2}({\Bbb U}_{2R})\rightarrow L_{m}^{2}({\Bbb U}_{2R}), \quad \tilde{Q}_{m, R} = {\rm I}_{L^{2}({\Bbb U}_{2R})}-\tilde{P}_{m, R}:L^{2}({\Bbb U}_{2R})\rightarrow L_{m}^{2}({\Bbb U}_{2R})^{\bot }, $

那么$ \tilde{P}_{m, R} $是$ m $ -维正交投影,且对于$ v\in L_{m}^{2}({\Bbb U}_{2R})^{\bot} $,有$ \mu _{m+1, R}\|\tilde{Q}_{m, R}v\|^{2}\leq\|\nabla v\|^{2}. $

对给定的$ m\in {\Bbb N}, $令

$ e_{m, R}(x) = \left\{ \begin{array}{ll} \tilde{e}_{m, R}(x), \quad &|x|<2R \\ 0, \quad \quad \quad \, \, \, \, \, &|x|\geq 2R\end{array} \right. , \quad m\in {\Bbb N}, $

那么$ \{e_{m, R}\}_{m\in {\Bbb N}} $是$ L^{2}({{\Bbb R}} ^{3}) $中的一族标准正交函数.令

$ L_{m, R}^{2}({{\Bbb R}} ^{3}) = {\rm span}\{e_{1, R}, e_{2, R}, \cdots , e_{m, R}\}, \quad L_{m, R}^{2}({{\Bbb R}} ^{3})^{\bot } = {\rm span} \{e_{m+1, R}, e_{m+2, R}, \cdots \}, $

$ \widehat{P}_{m, R}:L^{2}({{\Bbb R}} ^{3})\rightarrow L_{m, R}^{2}({{\Bbb R}} ^{3}), \quad \widehat{Q}_{m, R}:L^{2}({{\Bbb R}} ^{3})\rightarrow L_{m, R}^{2}({{\Bbb R}} ^{3})^{\bot }, $

那么$ \widehat{P}_{m, R} $是从$ L^{2}({{\Bbb R}} ^{3}) $到$ L_{m, R}^{2}({{\Bbb R}} ^{3}) $的$ m $ -维投影,且

$ \mu _{m+1, R}\|\widehat{Q}_{m, R}v\|^{2}\leq\|\nabla v\|^{2}\leq \|v\|_{1}^{2}, \quad \forall v\in L_{m, R}^{2}({{\Bbb R}} ^{3})^{\bot }. \label{3.8.3} $

令

$ {P}_{m, R}:L^{2}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3})\rightarrow \widehat{P}_{m, R}L^{2}({{\Bbb R}} ^{3})\times \widehat{P}_{m, R}L^{2}({\Bbb R}^{3}) = L_{m, R}^{2}({{\Bbb R}} ^{3})\times L_{m, R}^{2}({{\Bbb R}} ^{3}), \\ $

$ {Q}_{m, R}:L^{2}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3})\rightarrow \widehat{Q}_{m, R}L^{2}({{\Bbb R}} ^{3})\times \widehat{Q}_{m, R}L^{2}({\Bbb R}^{3}) = L_{m, R}^{2}({{\Bbb R}} ^{3})^{\bot }\times L_{m, R}^{2}({{\Bbb R}} ^{3})^{\bot }, $

那么$ P_{m, R} $是从$ L^{2}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3}) $到$ L_{m, R}^{2}({{\Bbb R}} ^{3})\times L_{m, R}^{2}({{\Bbb R}} ^{3}) $的$ 2m $ -维投影.令

$ y(r) = y(r, \tau - t, \theta _{-\tau }\omega , y_{\tau -t}(\theta _{-\tau }\omega )) = \psi_1(r) - \psi_2(r) = (y_1(r), \, y_2(r)) $

是系统(3.36)的解,其中$ y_{\tau -t}(\theta _{-\tau }\omega ) = \psi_{1, \tau -t}(\theta _{-\tau}\omega )-\psi_{2, \tau -t}(\theta _{-\tau}\omega ) $.设

$ y_{m, R}^{(1)}(r) = P_{m, R}y(r) = (\widehat{P}_{m, R}y_1(r), \widehat{P}_{m, R}y_2(r)) = (y^{(1)}_{1, m, R}(r), y^{(1)}_{2, m, R}(r)), $

$ y_{m, R}^{(2)}(r) = Q_{m, R}y(r) = (\widehat{Q}_{m, R}y_1(r), \widehat{Q}_{m, R}y_2(r)) = (y^{(2)}_{1, m, R}(r), y^{(2)}_{2, m, R}(r)), $

$ \begin{eqnarray*} y_{m, R}^{(3)}(r) & = & (I_{L^{2}({{\Bbb R}} ^{3})\times L^{2}({{\Bbb R}} ^{3})}-P_{m, R}-Q_{m, R})y(r) \\ & = & ((I_{L^{2}({{\Bbb R}} ^{3})}-\widehat{P}_{m, R}-\widehat{Q}_{m, R})y_1(r), (I_{L^{2}({{\Bbb R}} ^{3}) }-\widehat{P}_{m, R}-\widehat{Q} _{m, R})y_2(r)) \\ & = & (y^{(3)}_{1, m, R}(r), y^{(3)}_{2, m, R}(r)), \end{eqnarray*} $

其中

$ y_{i, m, R}^{(1)}(r) = \left\{ \begin{array}{ll} \tilde{P}_{m, R}y_i(r), \quad &|x|<2R \\ 0, \qquad\quad &|x|\geq 2R\end{array} \right. \in L_{m, R}^2(R^3), i = 1, 2, $

$ y_{i, m, R}^{(2)}(r) = \left\{ \begin{array}{ll} \tilde{Q}_{m, R}y_i(r), \quad &|x|<2R \\ 0, \qquad\quad &|x|\geq 2R \end{array} \right. \in L_{m, R}^2(R^3)^{\bot }, i = 1, 2, $

$ y^{(3)}_{i, m, R}(r) = \left\{ \begin{array}{ll} y_i(r), \quad &|x|\geq 2R, \\ 0, \quad &|x|<2R, \end{array} \quad i = 1, 2. \right. \label{3.69} $

那么系统(3.36)的解$ y(r) = (y_1(r), y_2(r)) $有如下分解

$ \left\{ \begin{array}{l} y_1(r) = y^{(1)}_{1, m, R}(r)+y^{(2)}_{1, m, R}(r)+y^{(3)}_{1, m, R}(r), \\ y_2(r) = y^{(1)}_{2, m, R}(r)+y^{(2)}_{2, m, R}(r)+y^{(3)}_{2, m, R}(r), \end{array} \quad r\geq \tau -t, \right. $

$ (y^{(1)}_{i, m, R}(r), y^{(2)} _{j, m, R}(r)) = (y^{(2)}_{i, m, R}(r), y^{(3)}_{j, m, R}(r)) = (y^{(1)}_{i, m, R}(r), y^{(3)}_{j, m, R}(r)) = 0, \quad i, j = 1, 2. $

接下来在$ E = L^2({{\Bbb R}} ^3)\times L^2({{\Bbb R}} ^3) $中估计$ y^{(2)}_{m, R}(r) = \big(y^{(2)}_{1, m, R}(r), \, y^{(2)}_{2, m, R}(r)\big) $和$ y^{(3)}_{m, R}(r) = \big(y^{(3)}_{1, m, R}(r), \, y^{(3)}_{2, m, R}(r)\big) $.

  引理3.4  对固定的$ R>0 $,取$ m\in {\Bbb N} $充分大使得$ 2\lambda +\mu _{m+1, R}>\frac{\sigma }{2} $,那么对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega, $存在随机变量$ {C}_{1}(\omega )\geq 0 $使得对任意$ \psi _{j, \tau -t}(\theta _{-\tau }\omega )\in {\chi}(\tau -t, \theta_{-t}\omega ) $, $ j = 1, 2 $和$ t\geq \frac{4}{\sigma }\ln (1+\frac{2\alpha \beta }{\sigma ^{2}}) $,有

(3.39) $ \begin{eqnarray} &&\|y_{m, R}^{(2)}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta _{-\tau }\omega ))\Vert _{E} {} \\ &\leq &\left( e^{\int_{-t}^{0}\left( bz(\theta _{s}\omega )-\frac{\sigma }{ 8}\right){\rm d}s} + \sqrt{\frac{\beta +\frac{\alpha \beta ^{2}}{\sigma ^{2}}}{ \sqrt{2\lambda +\mu _{m+1, R}}}}e^{\int_{-t}^{0}{C}_{1}(\theta _{s}\omega ){\rm d}s}\right) \|\psi _{1, \tau -t}(\theta _{-\tau }\omega ) - \psi _{2, \tau -t}(\theta _{-\tau }\omega )\|_{E}. {} \\ \end{eqnarray} $

证  取内积$ \big((3.36)_1 $, $ y_{1, m, R}^{(2)}\big) $有

(3.40) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\Vert y_{1, m, R}^{(2)}\Vert ^{2}+2\Vert \nabla y_{1, m, R}^{(2)}\Vert ^{2}+2(\lambda -bz(\theta _{r-\tau }\omega ))\Vert y_{1, m, R}^{(2)}\Vert ^{2}+2\alpha (y_{2}, \, y_{1, m, R}^{(2)}) {} \\ & = &2e^{-bz(\theta _{r-\tau }\omega )}\Big(f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(1)}(r))-f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(2)}(r)), \, y_{1, m, R}^{(2)}\Big). \end{eqnarray} $

根据(3.37)式和

$ \begin{eqnarray*} &&\|e^{-bz(\theta _{r-\tau }\omega )}[f(x, e^{bz(\theta _{r-\tau }\omega)}u_{1}^{(1)}(r))-f(x, e^{bz(\theta _{r-\tau }\omega)}u_{1}^{(2)}(r))]\|_{L^{\frac{6}{5}}({\Bbb U}_{2R})}^{2} \\&\leq &{{c}_{24}}e^{-2bz(\theta _{r-\tau }\omega )}\left( e^{4bz(\theta_{r-\tau }\omega )}L_{2}^{4}(\theta _{r-\tau }\omega )+1\right) \cdot\|y_{1}(r)\|^{2}, \end{eqnarray*} $

$ \begin{eqnarray*} &&2e^{-bz(\theta _{r-\tau }\omega )}\left( f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(1)}(r))-f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(2)}(r)), \, y_{1, m, R}^{(2)}\right) \\ &\leq &2\|e^{-bz(\theta _{r-\tau }\omega )}[f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(1)}(r))-f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(2)}(r))]\| _{L^{\frac{6}{5}}({\Bbb U}_{2R})}\| y_{1, m, R}^{(2)}\|_{L^{6}({\Bbb U}_{2R})} \\ &\leq &c_{25}\left( e^{2bz(\theta _{r-\tau }\omega )}L_{2}^{4}(\theta _{r-\tau }\omega )+e^{-2bz(\theta _{r-\tau }\omega )}\right) \| y_{1}(r)\|^{2}+\Vert \nabla y_{1, m, R}^{(2)}\|^{2}, \end{eqnarray*} $

$ 2\alpha (y_{2}, \, y_{1, m, R}^{(2)})\leq 2\sqrt{\frac{\alpha }{\beta }}\sqrt{\alpha }\|y_{2}\|\cdot \sqrt{\beta }\|y_{1, m, R}^{(2)}\|\leq \sqrt{\frac{\alpha }{\beta }}\|y\|_{E}^{2}, $

$ \mu _{m+1, R}\|y_{1, m, R}^{(2)}\|^{2}\leq \|\nabla y_{1, m, R}^{(2)}\|^{2}, $

(3.4)式可以变为

(3.41) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\Vert y_{1, m, R}^{(2)}\Vert ^{2}+\left( 2\lambda +\mu _{m+1, R}-2bz(\theta _{r-\tau }\omega )\right) \Vert y_{1, m, R}^{(2)}\Vert ^{2} {} \\ &\leq &\left( \sqrt{\frac{\alpha }{\beta }}+{c}_{26}e^{2bz(\theta _{r-\tau }\omega )}L_{2}^{4}(\theta _{r-\tau }\omega )+{c}_{26}e^{-2bz(\theta _{r-\tau }\omega )}\right) e^{\int_{\tau -t}^{r}2[bz(\theta _{s-\tau }\omega )+c_{2}]{\rm d}s}\|y_{\tau -t}(\theta _{-\tau }\omega )\|^{2}. {}\\ \end{eqnarray} $

对(3.41)式在$ [\tau -t, \tau ] $上应用Gronwall不等式,有

(3.42) $ \begin{eqnarray} &&\|y_{1, m, R}^{(2)}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta_{-\tau }\omega ))\|^{2} {} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{s}\omega )-2\lambda -\mu_{m+1, R}){\rm d}s}\|y_{1, m, R}^{(2)}(\tau -t)\|^{2} {} \\ &&+\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}\int_{\tau -t}^{\tau}\left( \sqrt{\frac{\alpha }{\beta }}+{c}_{26}e^{2bz(\theta _{s-\tau }\omega )}L_{2}^{4}(\theta _{s-\tau }\omega)+{c}_{26}e^{-2bz(\theta _{s-\tau}\omega )}\right) {} \\ && \times e^{\int_{\tau -t}^{s}2[bz(\theta _{r-\tau }\omega)+c_{2}]{\rm d}r}e^{\int_{s}^{\tau }(2bz(\theta _{r-\tau }\omega )-2\lambda -\mu_{m+1, R}){\rm d}r}{\rm d}s {} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{s}\omega )-2\lambda -\mu_{m+1, R}){\rm d}s}\|y_{1, m, R}^{(2)}(\tau -t)\|^{2} {} \\ &&+\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}e^{\int_{-t}^{0}2[bz(\theta _{r}\omega )+c_{2}]{\rm d}r} {} \\ && \times \int_{-t}^{0}\left( \sqrt{\frac{\alpha }{\beta }}+{c}_{26}e^{2bz(\theta _{s}\omega )}L_{2}^{4}(\theta _{s}\omega )+{c} _{26}e^{-2bz(\theta _{s}\omega )}\right) e^{(2\lambda +\mu _{m+1, R})s}{\rm d}s. \end{eqnarray} $

由$ \sqrt{x}\leq e^{x} $ ($ x\geq 0 $)和Hölder不等式可推出

(3.43) $ \begin{eqnarray} &&\int_{-t}^{0}\left( \sqrt{\frac{\alpha }{\beta }}+{c}_{26}e^{2bz(\theta _{s}\omega )}L_{2}^{4}(\theta _{s}\omega )+{c}_{26}e^{-2bz(\theta _{s}\omega )}\right) e^{(2\lambda +\mu _{m+1, R})s}{\rm d}s {} \\ &\leq &\frac{1}{\sqrt{2\lambda +\mu _{m+1, R}}}\, e^{\int_{-t}^{0}(\sqrt{\frac{\alpha }{\beta }}+{c}_{26}e^{2bz(\theta _{s}\omega )}L_{2}^{4}(\theta _{s}\omega )+{c}_{26}e^{-2bz(\theta _{s}\omega )})^{2}{\rm d}s}. \end{eqnarray} $

结合(3.42)和(3.43)式,可以得到

$ \begin{eqnarray*} &&\|y_{1, m, R}^{(2)}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta_{-\tau }\omega ))\|^{2} \\ &\leq &e^{\int_{-t}^{0}(2bz(\theta _{s}\omega )-2\lambda -\mu_{m+1, R}){\rm d}s}\|y_{1, m, R}^{(2)}(\tau -t)\|^{2} +\frac{1}{\sqrt{2\lambda +\mu_{m+1, R}}}e^{\int_{-t}^{0}2{C}_{1}(\theta _{s}\omega){\rm d}s}\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}, \end{eqnarray*} $

其中

$ {C}_{1}(\omega ) = b|z(\omega )|+c_{2}+\frac{1}{2}\left( \sqrt{\frac{\alpha}{\beta}}+{c}_{26}e^{2bz(\omega)}L_{2}^{4}(\omega )+{c}_{26}e^{-2bz(\omega)}\right) ^{2}. $

取内积$ \big((3.36)_2 $, $ y_{2, m, R}^{(2)}\big) $并由估计

$ 2\beta (y_{1}, \, y_{2, m, R}^{(2)}) = 2\beta(y_{1, m, R}^{(2)}, \, y_{2, m, R} ^{(2)})\leq \frac{\beta^{2}}{\sigma}\|y_{1, m, R} ^{(2)}\|^{2}+\sigma \|y_{2, m, R}^{(2)}\|^{2}, $

可以得到

(3.44) $ \begin{equation} \frac{\rm d}{{\rm d}t}\|y_{2, m, R}^{(2)}\|^{2}+[\sigma -2bz(\theta_{r-\tau }\omega)]\Vert y_{2, m, R}^{(2)}\|^{2}\leq \frac{\beta ^{2}}{\sigma}\|y_{1, m, R}^{(2)}\|^{2}. \end{equation} $

对(3.44)式在$ [\tau -t, \tau ] $上应用Gronwall不等式,有

$ \begin{eqnarray*} &&\|y_{2, m, R}^{(2)}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta _{-\tau }\omega ))\|^{2} \\ &\leq &e^{\int_{\tau -t}^{\tau }[2bz(\theta _{r-\tau }\omega )-\sigma ]{\rm d}r}\Vert y_{2, m, R}^{(2)}(\tau -t)\Vert ^{2} \\ &&+\frac{\beta ^{2}}{\sigma }\int_{\tau -t}^{\tau }e^{\int_{s}^{\tau }[2bz(\theta _{r-\tau }\omega )-\sigma ]{\rm d}r}\left( e^{\int_{-t}^{s-\tau }(2bz(\theta _{r}\omega )-2\lambda -\mu _{m+1, R}){\rm d}r}\|y_{1, m, R}^{(2)}(\tau -t)\|^{2}\right. \\ && + \frac{1}{\sqrt{2\lambda +\mu _{m+1, R}}}e^{\int_{-t}^{s-\tau }2{C}_{1}(\theta _{r}\omega ){\rm d}r}\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}\Big){\rm d}s \\ &\leq &e^{\int_{-t}^{0}[2bz(\theta _{r}\omega )-\sigma ]{\rm d}r}\Vert y_{2, m, R}^{(2)}(\tau -t)\Vert ^{2}+\frac{2\beta ^{2}}{\sigma ^{2}} e^{\int_{-t}^{0}[2bz(\theta _{r}\omega )-\frac{\sigma }{2}]{\rm d}r}\Vert y_{1, m, R}^{(2)}(\tau -t)\Vert ^{2} \\ &&+\frac{\beta ^{2}}{\sigma ^{2}\sqrt{2\lambda +\mu _{m+1, R}}} e^{\int_{-t}^{0}2{C}_{1}(\theta _{r}\omega ){\rm d}r}\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2} \\ &\leq &\left( \Vert y_{2, m, R}^{(2)}(\tau -t)\Vert ^{2}+\frac{2\beta ^{2}}{\sigma ^{2}}\Vert y_{1, m, R}^{(2)}(\tau -t)\Vert ^{2}\right) e^{\int_{-t}^{0}[2bz(\theta _{r}\omega )-\frac{\sigma }{2}]{\rm d}r} \\ &&+\frac{\beta ^{2}}{\sigma ^{2}\sqrt{2\lambda +\mu _{m+1, R}}}e^{\int_{-t}^{0}2{C}_{1}(\theta _{r}\omega ){\rm d}r}\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}. \end{eqnarray*} $

因此,

(3.45) $ \begin{eqnarray} &&\beta\|y_{1, m, R}^{(2)}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau-t}(\theta _{-\tau }\omega ))\Vert ^{2}+\alpha \|y_{2, m, R}^{(2)}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta _{-\tau }\omega ))\Vert ^{2}{} \\ &\leq &e^{\int_{-t}^{0}2bz(\theta _{s}\omega ){\rm d}s-\frac{\sigma }{2}t}\left( (1+\frac{2\alpha \beta }{\sigma ^{2}})\beta \Vert y_{1, m, R}^{(2)}(\tau-t)\Vert ^{2}+\alpha \Vert y_{2, m, R}^{(2)}(\tau -t)\Vert ^{2}\right){} \\ &&+\left( \beta +\frac{\alpha \beta ^{2}}{\sigma ^{2}}\right) \frac{1}{\sqrt{2\lambda +\mu _{m+1, R}}}e^{\int_{-t}^{0}2{C}_{1}(\theta _{s}\omega ){\rm d}s}\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2} {} \\ &\leq &(1+\frac{2\alpha \beta }{\sigma ^{2}})e^{-\frac{\sigma }{4} t}e^{\int_{-t}^{0}2bz(\theta _{s}\omega ){\rm d}s-\frac{\sigma }{4}t}\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}{} \\ &&+\left( \beta +\frac{\alpha \beta ^{2}}{\sigma ^{2}}\right) \frac{1}{\sqrt{ 2\lambda +\mu _{m+1, R}}}e^{\int_{-t}^{0}2{C}_{1}(\theta _{s}\omega){\rm d}s}\|y_{\tau -t}(\theta _{-\tau }\omega )\|_{E}^{2} {} \\ &\leq &e^{\int_{-t}^{0}2bz(\theta _{s}\omega ){\rm d}s-\frac{\sigma }{4}t}\|\psi _{1, \tau -t}(\theta _{-\tau }\omega )-\psi _{2, \tau -t}(\theta _{-\tau}\omega )\|_{E}^{2}\qquad \quad (t\geq \frac{4}{\sigma }\ln (1+\frac{2\alpha \beta}{\sigma ^{2}})){} \\ &&+\left( \beta + \frac{\alpha \beta ^{2}}{\sigma ^{2}}\right)\frac{1}{\sqrt{ 2\lambda +\mu _{m+1, R}}}e^{\int_{-t}^{0}2{C}_{1}(\theta _{s}\omega){\rm d}s}\|\psi _{1, \tau -t}(\theta _{-\tau }\omega ) - \psi _{2, \tau -t}(\theta _{-\tau }\omega )\|_{E}^{2}, \end{eqnarray} $

引理3.4证毕.

引理3.5  对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega , $$ t\geq 0, $$ m\in {\Bbb N} $和$ R\geq 2, $存在随机变量$ {C}_{2}(\omega )\geq 0 $,使得对任意的$ \psi _{j, \tau -t}(\theta _{-\tau }\omega )\in{\chi}(\tau -t, \theta _{-t}\omega ) $, $ j = 1, $$ 2, $下式成立

(3.46) $ \begin{eqnarray} &&\|y_{m, R}^{(3)}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta_{-\tau }\omega ))\|_{E} {} \\ &\leq &\left( e^{\int_{-t}^{0}\left( bz(\theta _{r}\omega )-\frac{{\rho }}{4}\right){\rm d}r}+\sqrt{\left( \frac{1}{\sqrt{{\rho }}}+\frac{1}{\sqrt[4]{{\rho }} \sqrt{\beta }}\right) J_{\varepsilon , R}}\, e^{\int_{-t}^{0}{C}_{2}(\theta _{r}\omega ){\rm d}r}\right) {} \\ && \times \Vert \psi _{1, \tau -t}(\theta _{-\tau }\omega )-\psi _{2, \tau -t}(\theta _{-\tau }\omega )\Vert _{E}, \end{eqnarray} $

其中

$ J_{\varepsilon , R} = \varepsilon +\frac{1}{R^{2}}+\frac{1}{R}+\zeta _{4, \frac{R}{2}}+\beta _{4, R}, \quad \beta _{4, R} = \left( {\int_{|x|\geq R}}\beta _{4}^{3}(x){\rm d}x\right) ^{\frac{1}{3}}. $

证  取内积$ \Big((3.36)_1 $, $ \beta \vartheta (\frac{|x|^{2}}{R^{2}})y_{1}\Big) $和$ \Big((3.36)_2 $, $ \alpha \vartheta (\frac{|x|^{2}}{R^{2}})y_{2}\Big) $并相加,对于$ R\geq2 $有

(3.47) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\left( \beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|y_{1}|^{2}{\rm d}x+\alpha \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|y_{2}|^{2}{\rm d}x\right) -2\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{ |x|^{2}}{R^{2}})\Delta y_{1}y_{1}{\rm d}x {} \\ && +2({\rho}-bz(\theta _{r-\tau }\omega ))\left( \beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|y_{1}|^{2}{\rm d}x+\alpha \int_{ {{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|y_{2}|^{2}{\rm d}x\right) {} \\ & = &2\beta e^{-bz(\theta _{r-\tau }\omega )}\int_{{{\Bbb R}} ^{3}}\vartheta ( \frac{|x|^{2}}{R^{2}})[f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(1)}(r)) - f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(2)}(r))]y_{1}{\rm d}x. \end{eqnarray} $

由(3.1)式和$ \vartheta $的定义可以推出,对任意的$ u\in H^{1}({{\Bbb R}} ^{3}) $有

(3.48) $ \begin{eqnarray} \Vert u\Vert _{L^{6}({{\Bbb R}} ^{3}\backslash {\Bbb U}_{2R})}^{2} & = &\left( {\int_{|x|\geq 2R}}u^{6}{\rm d}x\right) ^{\frac{2}{6}}\leq \left( {\int_{ {{\Bbb R}} ^{3}}}\left( \vartheta (\frac{|x|^{2}}{R^{2}})u\right) ^{6}{\rm d}x\right) ^{\frac{2}{6}} {} \\ &\leq &C_{0}^{2}\left( {\int_{{{\Bbb R}} ^{3}}}\left( \nabla \left( \vartheta (\frac{|x|^{2}}{R^{2}})u\right) \right) ^{2}{\rm d}x+{\int_{{{\Bbb R}} ^{3}}} \vartheta (\frac{|x|^{2}}{R^{2}})u^{2}{\rm d}x\right) {} \\ &\leq &\frac{16C_{0}^{2}\tilde{C}^{2}}{R^{2}}\Vert u\Vert ^{2}+2C_{0}^{2}{ \int_{{{\Bbb R}} ^{3}}}\vartheta (\frac{|x|^{2}}{R^{2}})[\left( \nabla u\right) ^{2}+u^{2}]{\rm d}x. \end{eqnarray} $

应用(3.1), (3.35)和(3.48)式可知,对$ j = 1 $, $ 2 $和$ R\geq 2 $有

(3.49) $ \begin{eqnarray} \Vert u_{1}^{(j)}(r)\Vert _{L^{6}({{\Bbb R}} ^{3}\backslash {\Bbb U} _{R})}^{2} &\leq &\frac{64\beta ^{-1}C_{0}^{2}\tilde{C}^{2}}{R^{2}} L_{2}^{2}(\theta _{r-\tau }\omega ) {} \\ &&+2C_{0}^{2}\bar{\beta}^{-2}\left( \varepsilon +\frac{2Q_{1}(\theta _{r-\tau }\omega )+2Q_{2}(\theta _{r-\tau }\omega )}{R}+\zeta _{4, \frac{R}{2} }Q_{0}(\theta _{r-\tau }\omega )\right) {} \\ &\leq &2C_{0}^{2}\bar{\beta}^{-2}\varepsilon +Q_{4}(\theta _{r-\tau }\omega )\left( \frac{1}{R^{2}}+\frac{1}{R}+\zeta _{4, \frac{R}{2}}\right) , \end{eqnarray} $

其中

$ \begin{eqnarray*} Q_{4}(\omega ) = 64\beta ^{-1}C_{0}^{2}\tilde{C}^{2}L_{2}^{2}(\omega )+2C_{0}^{2}\bar{\beta}^{-2}\left( Q_{0}(\omega )+2Q_{1}(\omega )+2Q_{2}(\omega )\right) . \end{eqnarray*} $

由(1.2), (3.1), (3.49)式和Hölder不等式,有

(3.50) $ \begin{eqnarray} &&\int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})[f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(1)}(r))-f(x, e^{bz(\theta _{r-\tau }\omega )}u_{1}^{(2)}(r))]y_{1}{\rm d}x {} \\ &\leq &\left[ 6c_{3}C_{0}e^{2bz(\theta _{r-\tau }\omega )}\left( \Vert u_{1}^{(1)}(r)\Vert_{L^{6}({{\Bbb R}} ^{3}\backslash {\Bbb U} _{R})}^{2}+\Vert u_{1}^{(2)}(r)\Vert _{L^{6}({{\Bbb R}} ^{3}\backslash {\Bbb U}_{R})}^{2}\right) +2C_{0}\beta _{4, R}\right] \Vert y_{1}\Vert _{1}\cdot \Vert y_{1}\Vert {} \\ &\leq &\left[ 12c_{3}C_{0}e^{2bz(\theta _{r-\tau }\omega )}\left( 2C_{0}^{2} \bar{\beta}^{-2}\varepsilon +Q_{4}(\theta _{r-\tau }\omega )(\frac{1}{R^{2}}+ \frac{1}{R}+\zeta _{4, \frac{R}{2}})\right) +2C_{0}\beta _{4, R}\right]{} \\ && \times \left(\|y_{1}\|^{2}+\|y_{1}\|\cdot\|\nabla y_{1}\|\right). \end{eqnarray} $

根据(3.37)式可知

(3.51) $ \begin{eqnarray} \|y_{1}(r)\|^{2} + \|y_{1}(r)\|\cdot\|\nabla y_{1}(r)\| &\leq &\beta ^{-1}e^{\int_{\tau -t}^{r}2[bz(\theta _{s-\tau }\omega )+c_{2}]{\rm d}s}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2} {} \\ && +\frac{1}{\sqrt{{\beta}}}e^{\int_{\tau -t}^{r}[bz(\theta _{s-\tau }\omega ) +c_{2}]{\rm d}s}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E} \|\nabla y_{1}(r)\|.{\qquad} \end{eqnarray} $

应用(3.50)式, (3.51)式和$ 2\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})\Delta y_{1}y_{1}{\rm d}x\leq \frac{4\sqrt{2}\tilde{C}\beta}{R}\|y_{1}\|\cdot\|\nabla y_{1}\| $,可以将(3.47)式变为

(3.52) $ \begin{eqnarray} &&\frac{\rm d}{{\rm d}t}\left( \beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{ R^{2}})|y_{1}|^{2}{\rm d}x+\alpha \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{ R^{2}})|y_{2}|^{2}{\rm d}x\right) {} \\ &&+2({\rho }-bz(\theta _{r-\tau }\omega ))\left( \beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|y_{1}|^{2}{\rm d}x+\alpha \int_{ {{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|y_{2}|^{2}{\rm d}x\right) {} \\ &\leq &J_{\varepsilon , R}\, \Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}e^{\int_{\tau -t}^{r}2[bz(\theta _{s-\tau }\omega )+c_{2}]{\rm d}s}Q_{5}(\theta _{r-\tau }\omega ) {} \\ && +J_{\varepsilon , R}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}\, e^{\int_{\tau -t}^{r}[bz(\theta _{l-\tau }\omega )+c_{2}]{\rm d}l}\|\nabla y_{1}(r)\|Q_{5}(\theta _{r-\tau }\omega ), \end{eqnarray} $

其中$ Q_{5}(\omega ) = {c}_{27}e^{b|z(\omega )|}\left( 1+L_{2}^{2}(\omega )+Q_{0}(\omega )+Q_{1}(\omega )+Q_{2}(\omega )\right) $.对(3.52)式在$ [\tau -t, \tau ] $ ($ t\geq 0 $)上应用Gronwall不等式可得

$ \begin{eqnarray*} &&\beta \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}})|y_{1}(\tau )|^{2}{\rm d}x+\alpha \int_{{{\Bbb R}} ^{3}}\vartheta (\frac{|x|^{2}}{R^{2}} )|y_{2}(\tau )|^{2}{\rm d}x \\ &\leq &e^{\int_{-t}^{0}\left( 2bz(\theta _{r}\omega )-2{\rho }\right) {\rm d}r}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2} \\ &&+J_{\varepsilon , R}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}\int_{\tau -t}^{\tau }Q_{5}(\theta _{s-\tau }\omega )e^{\int_{\tau -t}^{s}2(bz(\theta _{r-\tau }\omega )+c_{2}){\rm d}r}e^{\int_{s}^{\tau }\left( 2bz(\theta _{r-\tau }\omega )-2{\rho }\right){\rm d}r}{\rm d}s \\ &&+J_{\varepsilon , R}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}\int_{\tau -t}^{\tau }\|\nabla y_{1}(s)\|Q_{5}(\theta _{s-\tau }\omega )e^{\int_{\tau -t}^{s}(bz(\theta _{r-\tau }\omega )+c_{2}){\rm d}r}e^{\int_{s}^{\tau }\left( 2bz(\theta _{r-\tau }\omega )-2{\rho } \right){\rm d}r}{\rm d}s \\ &\leq &e^{\int_{-t}^{0}\left( 2bz(\theta _{r}\omega )-2{\rho }\right) {\rm d}r}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}\\ && + J_{\varepsilon , R}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}e^{2\int_{-t}^{0}(bz(\theta _{r}\omega )+c_{2}){\rm d}r} \int_{-t}^{0} Q_{5}(\theta _{s}\omega )e^{2{\rho }s}{\rm d}s \\ &&+J_{\varepsilon , R}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}e^{\int_{-t}^{0}(b|z(\theta _{r}\omega )|+c_{2}){\rm d}r}\left( \int_{\tau -t}^{\tau }\left( e^{\int_{s}^{\tau }[bz(\theta _{r-\tau }\omega )+c_{2}]{\rm d}r}\Vert \nabla y_{1}(s)\Vert \right) ^{2}{\rm d}s\right) ^{\frac{1}{2}} \\ &&\times \left( \int_{-t}^{0}(Q_{5}(\theta _{s}\omega )e^{2{\rho } s})^{2}{\rm d}s\right) ^{\frac{1}{2}} \\ &\leq &\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}\left( e^{\int_{-t}^{0}\left( 2bz(\theta _{r}\omega )-2{\rho }\right){\rm d}r}+\frac{1}{2 \sqrt{{\rho }}}J_{\varepsilon , R}e^{\int_{-t}^{0}[2(b|z(\theta _{r}\omega )|+c_{2})+Q_{5}^{2}(\theta _{r}\omega )]{\rm d}r}\right) \\ &&+\frac{1}{\sqrt[4]{8{\rho }}\sqrt{2\beta }}J_{\varepsilon , R}\Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}e^{\int_{-t}^{0}[2(b|z(\theta _{r}\omega )|+c_{2})+\frac{1}{2}Q_{5}^{4}(\theta _{r}\omega )]{\rm d}r} \\ &\leq &\left( e^{\int_{-t}^{0}\left( 2bz(\theta _{r}\omega )-2{\rho }\right) {\rm d}r} + \left( \frac{1}{\sqrt{{\rho }}} + \frac{1}{\sqrt[4]{{\rho }}\sqrt{\beta }} \right) J_{\varepsilon , R}e^{\int_{-t}^{0}[2(b|z(\theta _{r}\omega )|+c_{2})+1+Q_{5}^{4}(\theta _{r}\omega )]{\rm d}r}\right) \Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2} \\ &\leq &\left( e^{\int_{-t}^{0}\left( 2bz(\theta _{r}\omega )-\frac{{\rho }}{2 }\right){\rm d}r} + \left( \frac{1}{\sqrt{{\rho }}} + \frac{1}{\sqrt[4]{{\rho }}\sqrt{ \beta }}\right) J_{\varepsilon , R}e^{\int_{-t}^{0}[2(b|z(\theta _{r}\omega )|+c_{2})+1+Q_{5}^{4}(\theta _{r}\omega )]{\rm d}r}\right) \Vert y_{\tau -t}(\theta _{-\tau }\omega )\Vert _{E}^{2}, \end{eqnarray*} $

其中

$ {C}_{2}(\omega ) = b|z(\omega )|+c_{2}+1+Q_{5}^{4}(\omega ). $

因此, (3.46)式成立.引理3.5证毕.

引理3.6  对固定的$ R\geq 2 $,令$ m\in {\Bbb N} $足够大使得$ 2\lambda+\mu _{m+1, R}>\frac{\sigma }{2} $,那么对任意$ \tau \in {{\Bbb R}} $, $ \omega \in \Omega $,存在随机变量$ {C}_{3}(\omega )\geq 0 $和$ 2m $ -维正交投影$ P_{m, R} $: $ L^{2}({{\Bbb R}} ^{3})\times L^{2}({\Bbb R}^{3})\rightarrow L_{m, R}^{2}({{\Bbb R}} ^{3})\times L_{m, R}^{2}({{\Bbb R}} ^{3}) $使得对任意的$ \psi _{j, \tau -t}(\theta _{-\tau }\omega ) $$ \in $$ {\cal {\chi }}(\tau -t, \theta _{-t}\omega ) $, $ j = 1 $, $ 2 $,有

(ⅰ)对任意$ t\geq 0 $,

(3.53) $ \begin{eqnarray} &&\|\Psi (t, \tau -t, \theta _{-t}\omega )\psi _{1, \tau -t}(\theta _{-\tau }\omega )-\Psi (t, \tau -t, \theta _{-t}\omega )\psi _{2, \tau -t}(\theta _{-\tau }\omega )\|_{E} {} \\ &\leq &e^{\int_{-t}^{0}{C}_{3}(\theta _{s}\omega ){\rm d}s}\Vert \psi _{1, \tau -t}(\theta _{-\tau }\omega )-\psi _{2, \tau -t}(\theta _{-\tau }\omega )\Vert _{E}. \end{eqnarray} $

(ⅱ)对任意$ t\geq \max\{\frac{8\ln2}{ {\rho}}, \frac{4}{\sigma} \ln(1+\frac{2\alpha \beta}{\sigma^2})\} $,

(3.54) $ \begin{eqnarray} &&\|(I - P_{m, R})\Psi (t, \tau - t, \theta _{-t }\omega )\psi_{1, \tau -t}(\theta _{-\tau}\omega ) - (I - P_{m, R})\Psi (t, \tau - t, \theta _{-t}\omega )\psi_{2, \tau -t}(\theta _{-\tau}\omega )\|_E {} \\ \ &\leq &\|y^{(2)}_{m, R}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta _{-\tau }\omega ))\Vert_E +\|y^{(3)}_{m, R}(\tau , \tau -t, \theta _{-\tau }\omega , y_{\tau -t}(\theta _{-\tau }\omega ))\Vert_{E} {} \\ &\leq &\left( e^{\int_{-t}^{0}\left( b |z(\theta _{s }\omega )|-\frac{ {\rho} }{8 }\right){\rm d}s}+\frac{1}{2} \widehat{\delta} _{\varepsilon , R, m}e^{\int_{-t}^{0} {C}_{3}(\theta _{s}\omega ){\rm d}s}\right) \Vert \psi_{1, \tau -t}(\theta _{-\tau }\omega )-\psi_{2, \tau -t}(\theta _{-\tau }\omega )\Vert_E , \end{eqnarray} $

其中

$ \frac{1}{2} \widehat{\delta }_{\varepsilon, R, m} = \sqrt{\left( \frac{1}{\sqrt{ {\rho }}}+\frac{1}{ \sqrt[4]{ {\rho} }\sqrt{\beta}}\right) J_{\varepsilon , R}} +\sqrt{\frac{\beta+\frac{\alpha \beta^2}{\sigma^2}}{\sqrt {2\lambda+ \mu_{m+1, R}}}}\, , $

(3.55) $ \begin{eqnarray} {C}_{3}(\omega )& = &{c}_{28}+3b |z(\omega )|+ {c}_{29}e^{4b| z(\omega )|}L_{2}^{8}(\omega ) +{\ {c}_{30}}e^{4b |z(\omega )|} {} \\ &&+ {c}_{31}e^{4b| z(\omega )|}\left( Q_{0}^{4}(\omega )+Q_{1}^{4}(\omega )+Q_{2}^{4}(\omega )\right). \end{eqnarray} $

证  应用(3.38), (3.39)和(3.46)式,可得

$ b|z(\omega )|+c_{2}+{C}_{1}(\omega )+{C}_{2}(\omega )<{C}_{3}(\omega ), $

对任意$ t\geq \frac{8\ln 2}{{\rho }} $,有

$ e^{\int_{-t}^{0}\left( bz(\theta _{s}\omega )-\frac{{\rho }}{4}\right) {\rm d}s}\leq \frac{1}{2}\, e^{\int_{-t}^{0}\left( bz(\theta _{s}\omega )-\frac{{\rho }}{8}\right){\rm d}s}, $

$ e^{\int_{-t}^{0}\left( bz(\theta _{s}\omega )-\frac{\sigma }{8}\right) {\rm d}s}\leq e^{\int_{-t}^{0}\left( bz(\theta _{s}\omega )-\frac{{\rho }}{4}\right){\rm d}s}. $

再次应用(3.38), (3.39)和(3.46)式,可得(ⅰ)和(ⅱ)成立.引理3.6证毕.

接下来验证期望$ {\bf E}[b|z(\omega )|] $, $ {\bf E}[{C}_{3}(\omega )] $和$ {\bf E}[{C}_{3}^{2}(\omega )] $的有界性.首先介绍下面的引理.

引理3.7^[45]  Ornstein-Uhlenbeck过程$ z(\theta _{t}\omega ) $满足

$ {\bf E}\left[ e^{\eta \int_{\tau }^{\tau +t}|z(\theta _{s}\omega )|{\rm d}s}\right] \leq e^{\eta t}\, \, \, \rm{for }\, \, 0\leq \eta ^{2}\leq 1, \mathit{\ }\tau \in {{\Bbb R}} , \quad t\geq 0, \label{3.13.1} $

(3.56) $ \begin{equation} {\bf E}[|z(\theta _{t}\omega )|^{p}] = \frac{\Gamma (\frac{1+p}{2})}{\sqrt{\pi}}, \quad \forall p>0, \quad t\in {{\Bbb R}} , \end{equation} $

其中$ \Gamma $是Gamma函数.

引理3.8  若系统(1.1)中的参数$ b $满足

(3.57) $ \begin{equation} b<\min \left\{ \frac{{\rho }\sqrt{\pi }}{64}, \frac{\rho}{6144}, \frac{1}{3072}\right\} , \end{equation} $

那么

$ 0\leq {\bf E}[b|z(\omega )|]\leq \frac{{\rho }}{64}, \quad {0\leq }{\bf E}[{C}_{3}(\omega )], \, {\bf E}[{C}_{3}^{2}(\omega )]<\infty . $

证  显然,由(3.56)式, (3.57)式和$ \Gamma (1) = 1 $可以推出$ 0\leq {\bf E}[b|z(\omega )|]\leq \frac{{\rho }}{64} $.应用(3.34)和(3.55)式有

(3.58) $ \begin{equation} C^2_3(\omega)\leq c_{32}(1+|z(\omega)|^2+e^{16b|z(\omega)|} +L_2^{32}(\omega)+Q_0^{16}(\omega)+Q_1^{16}(\omega)+ Q_3^{16}(\omega)). \end{equation} $

根据(3.4)和(3.5)式可得

(3.59) $ \begin{equation} L_{1}^{32}(\omega) \leq c_{33}\left(1+\left(\int_{-1}^{\, 0}e^{-2bz(\theta_{s}\omega)}{\rm d}s \right)^{16} +\max\limits_{-1\leq s\leq 0}|z(\theta_{s}\omega)|^{32}+ \tilde{Q}_1^{64}(\omega)\right), \end{equation} $

(3.60) $ \begin{equation} \tilde{Q}_1^{64}(\omega)\leq c_{34}(1+e^{\int_{-1}^{\, 0} {128bz(\theta_{l}\omega)}{\rm d}l}+ {Q}_0^{64}(\omega)). \end{equation} $

由(3.3), (3.6), (3.58), (3.59)和(3.60)式,有

(3.61) $ \begin{eqnarray} C^2_3(\omega)&\leq & c_{35}\bigg(1+|z(\omega)|^2+e^{16b|z(\omega)|} + Q_6^{16}(\omega)+Q_7^{16}(\omega)+ Q_0^{64}(\omega)+ \left(\int_{-1}^{\, 0}e^{-2bz(\theta_{s}\omega)}{\rm d}s \right)^{16}{} \\ &&\left.+\max\limits_{-1\leq s\leq 0}|z(\theta_{s}\omega)|^{32}+e^{\int_{-1}^{\, 0} {128bz(\theta_{l}\omega)}{\rm d}l} \right), \end{eqnarray} $

其中

$ {Q}_6(\omega) = \int_{-\infty }^{0}e^{\int_{s}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho}}{2}){\rm d}l}L_{2}^{2}(\theta_{s}\omega){\rm d}s, $

$ Q_{7}(\omega) = \int_{-\infty}^{0}e^{\int_{s}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho}}{2}){\rm d}l}\left(z^{2}(\theta _{s}\omega)L_{2}^{2}(\theta_{s}\omega)+e^{4bz(\theta_{s}\omega )}L_{2}^{6}(\theta_{s}\omega )\right){\rm d}s. $

令

(3.62) $ \begin{equation} b< \min\bigg\{\frac{\rho}{32n}, \frac{1}{16n}\bigg\}, \quad n\in {\Bbb N}. \end{equation} $

注意到$ L_0^{2n}(\omega)\leq(1+c_4)^n2^{n-1}(1+Q_0^n(\omega)). $应用Hölder不等式可得

(3.63) $ \begin{eqnarray} {\bf E}[{Q}_0^{n}(\omega)]&\leq& {\bf E}\left( \int_{-\infty }^{0}e^{\, \frac{(n-1)\rho}{2n}s} e^{\, \frac{\rho}{2n}s-2bz(\theta_{s}\omega)+ \int_{s}^{0}2bz(\theta_{l}\omega){\rm d}l}{\rm d}s\right)^{n}{} \\ & \leq&(\frac{2}{\rho})^{n-1}{\bf E}\int_{-\infty }^{0}e^{\, \frac{\rho}{2}s} e^{\, 2bn|z(\theta_{s}\omega)|} e^{\int_{s}^{0}2bnz(\theta_{l}\omega){\rm d}l}{\rm d}s{} \\ & \leq&(\frac{2}{\rho})^{n-1}\int_{-\infty }^{0}e^{\, \frac{\rho}{2}s}\Big( {\bf E}\big[e^{\, 4bn|z(\theta_{s}\omega)|}\big]+ {\bf E}\big[e^{\int_{s}^{0}4bnz(\theta_{l}\omega){\rm d}l} \big]\Big){\rm d}s{}\\ &\doteq&\kappa_{0, n}. \end{eqnarray} $

从(3.62)式,引理3.7和

(3.64) $ \begin{equation} {\bf E}[e^{|\epsilon z(\theta_{s}\omega)|}]\leq \left( 1+\frac{|\epsilon|}{\sqrt{\pi}}\right) e^{|\epsilon|^2}, \, \, \, \, \forall \epsilon, \, s\in {{\Bbb R}} , \qquad( \rm{见文献[28]的(3.68)式)} \end{equation} $

可以推出

$ {\bf E}[L_{0}^{2n}(\theta_s\omega)] = {\bf E}[L_{0}^{2n}(\omega)]\leq (1+c_4)^n2^{n-1}(1+\kappa_{0, n})\doteq \xi_{0, n}<\infty, \quad s\in{{\Bbb R}} . $

在(3.63)式中用$ 4n $替换$ n $,得到

$ {\bf E}[{Q}_0^{4n}(\omega)]\leq (\frac{2}{\rho})^{4n-1}\int_{-\infty }^{0}e^{\, \frac{\rho}{2}s}\Big( {\bf E}\big[e^{\, 16bn|z(\theta_{s}\omega)|}\big]+ {\bf E}\big[e^{\int_{s}^{0}16bnz(\theta_{l}\omega){\rm d}l} \big]\Big){\rm d}s\doteq\kappa_{0, 4n}. $

考虑到(3.62)式, (3.64)式和引理3.7,从(3.4)式可知,对$ \forall s\in{{\Bbb R}} $有

(3.65) $ \begin{eqnarray} &&{\bf E}[L_{1}^{2n}(\theta_s\omega)] = {\bf E}[L_{1}^{2n}(\omega)]{} \\ &\leq& c_{36}5^{n-1}{\bf E}\left(1 + \left(\int_{-1}^{\, 0}e^{-2bz(\theta_{s}\omega)}{\rm d}s \right)^{n}+\max\limits_{-1\leq s\leq 0}|z(\theta_{s}\omega)|^{2n}+e^{\int_{-1}^08bn|z(\theta_l\omega)|{\rm d}l}+{Q}_0^{4n}(\omega) \right){} \\ &\leq& c_{36}5^{n-1}\left(1 + \int_{-1}^{\, 0} {\bf E}\big[e^{2bn|z(\theta_{s}\omega)|}\big]{\rm d}s+ {\bf E}\big[|z(\theta_{s}\omega)|^{2n}\big]+ {\bf E}\big[e^{\int_{-1}^08bn|z(\theta_l\omega)|{\rm d}l}\big] +{\bf E}\big[{Q}_0^{4n}(\omega)\big]\right){} \\ &\doteq&\xi_{1, n}<\infty. \end{eqnarray} $

因此

(3.66) $ \begin{equation} {\bf E}[L_{2}^{2n}(\theta_s\omega)] = {\bf E}[L_{2}^{2n}(\omega)] \leq2^{n-1}(\xi_{0, n}+\xi_{1, n})\doteq\xi_{2, n}<\infty, \quad s\in{{\Bbb R}} . \end{equation} $

由Hölder不等式, (3.57)式, (3.66)式和引理3.7可以得到

(3.67) $ \begin{eqnarray} {\bf E}[Q_6^{16}(\omega)]& = &{\bf E}\left(\int_{-\infty}^0 e^{\frac{15}{32}\rho s}e^{\int_s^0(2bz(\theta_l\omega)- \frac{\rho}{32}){\rm d}l}L_2^2(\theta_s\omega){\rm d}s\right)^{16}{} \\&\leq& (\frac{2}{\rho})^{15}\int_{-\infty}^0e^{\frac{\rho s}{2}}\Big( {\bf E}[e^{\int_s^064bz(\theta_l\omega){\rm d}l}]+ {\bf E}[L_{2}^{64}(\theta_s\omega)]\Big){\rm d}s\doteq \xi_3<\infty. \end{eqnarray} $

应用Hölder不等式,有

$ \begin{eqnarray*} Q_{7}^{16}(\omega)& = &\left(\int_{-\infty}^{0}e^{\int_{s}^{0}(2bz(\theta _{l}\omega )-\frac{{\rho}}{2}){\rm d}l}\left(z^{2}(\theta _{s}\omega)L_{2}^{2}(\theta_{s}\omega)+e^{4bz(\theta_{s}\omega )}L_{2}^{6}(\theta_{s}\omega )\right){\rm d}s\right)^{16}\\ & \leq& (\frac{2}{\rho})^{15} \int_{-\infty}^{0}e^{\int_{s}^{0}(32bz(\theta _{l}\omega )-\frac{{\rho}}{2}){\rm d}l}\left(z^{32}(\theta _{s}\omega)L_{2}^{32}(\theta_{s}\omega)+e^{64bz(\theta_{s}\omega )}L_{2}^{96}(\theta_{s}\omega )\right){\rm d}s. \end{eqnarray*} $

由Hölder不等式, (3.57)式, (3.66)式和引理3.7,可以推出

$ \begin{eqnarray*} &&{\bf E}\int_{-\infty}^{0}e^{\int_{s}^{0}(32bz(\theta _{l}\omega )-\frac{{\rho}}{2}){\rm d}l} z^{32}(\theta _{s}\omega)L_{2}^{32}(\theta_{s}\omega){\nonumber} \\ & \leq& \int_{-\infty}^{0}e^{\frac{\rho}{2}s}\Big( {\bf E}\big[e^{\int_{s}^{0}64bz(\theta_{l}\omega){\rm d}l}\big] +{\bf E}[z^{128}(\theta_{s}\omega)]+ {\bf E}[L_{2}^{128}(\theta_{s}\omega)]\Big){\rm d}s\doteq\xi_4 <\infty, \\ &&{\bf E}\int_{-\infty}^{0}e^{\int_{s}^{0}(32bz(\theta _{l}\omega )-\frac{{\rho}}{2}){\rm d}l} e^{64bz(\theta _{s}\omega)}L_{2}^{96}(\theta_{s}\omega){\rm d}s{\nonumber} \\ & \leq& \int_{-\infty}^{0}e^{\frac{\rho}{2}s}\Big( {\bf E}\big[e^{\int_{s}^{0}64bz(\theta_{l}\omega){\rm d}l}\big] +{\bf E}[z^{256}(\theta_{s}\omega)]+ {\bf E}[L_{2}^{384}(\theta_{s}\omega)]\Big){\rm d}s\doteq\xi_5 <\infty. \end{eqnarray*} $

从而

(3.68) $ \begin{equation} {\bf E}[Q_{7}^{16}(\omega)]\leq (\frac{2}{\rho})^{15}(\xi_4+\xi_5)\doteq\xi_6<\infty. \end{equation} $

根据引理3.7, (3.57), (3.61), (3.63), (3.64), (3.67)和(3.68)式可以得到

$ {\bf E}[C_{3}^{2}(\omega)]\leq\xi_7<\infty, \quad {\bf E}[C_{3}(\omega)] \leq\big({\bf E}[C_{3}^{2}(\omega)]\big)^{1/2} <\infty. $

引理3.8证毕.

下面给出本节的主要结果.

定理3.1  假设(A1)–(A2)和(3.57)式成立.那么$ \{\Psi (t, \tau , \omega )\}_{t\geq 0, \tau \in {{\Bbb R}} , \omega \in \Omega } $存在一个随机指数吸引子$ \{{\cal O}(\tau , \omega )\}_{\tau \in {{\Bbb R}} , \omega \in \Omega } $,且有如下性质:对任意$ \tau \in {{\Bbb R}} $和$ \omega \in \Omega $,

(ⅰ) $ {\cal O}(\tau , \omega )\, (\subseteq \overline{\chi(\tau, \omega )}) $是$ E $的紧集且关于$ \omega $可测;

(ⅱ) $ \Psi(t, \tau , \omega){\cal O}(\tau , \omega )\subseteq {\cal O}(t+\tau , \theta _{t}\omega ) $, $ \forall t\geq 0 $;

(ⅲ)存在正数$ l_{0}\in {\Bbb N} $使得$ \dim _{f}{\cal O}(\tau, \omega)\leq l_{0}<\infty $;

(ⅳ)对任意$ D\in {\cal D}(E), $存在随机变量$ \check{T}_{D}(\tau , \omega )\geq 0 $和缓增随机变量$ \check{b}_{\omega }>0 $使得

$ {\rm d}_{h}(\Psi (t, \tau -t, \theta _{-t}\omega )D(\tau -t, \theta _{-t}\omega ), {\cal O}(\tau , \omega ))\leq \check{b}_{\omega }e^{-\frac{ \ln \frac{4}{3}}{\max \{\frac{128\ln 2}{\rho }, \, \frac{16}{\sigma }\ln (1+ \frac{2\alpha \beta }{\sigma ^{2}})\}}t}, \, \, t\geq \check{T}_{D}(\tau , \omega ). \label{4.31.0} $

证  在(3.53)式的右端和(3.54)式中取

$ \frac{4\ln 2}{\frac{\rho }{8}}\leq t = t_{0} = \max \left\{ \frac{32\ln 2}{\rho} , \, \frac{4}{\sigma}\ln (1+\frac{2\alpha \beta }{\sigma ^{2}})\right\} <\infty, $

从引理3.8可得

$ \xi_{7} = t_{0}^{2}\left( {\bf E}[{C}_{3}^{2}(\omega )]+\frac{{\rho }}{8} {\bf E}[{C}_{3}(\omega )]\right)<\infty, $

记

$ \eta = \min \left\{ \frac{1}{4}e^{-\frac{\rho t_{0}}{16}}, \, \, e^{-\frac{\xi_{7}}{\ln \frac{3}{2}}}\right\}. $

由于

$ J_{\varepsilon , R} = \varepsilon +\frac{1}{R^{2}}+\frac{1}{R}+\zeta _{4, \frac{R }{2}}+\beta _{4, R}, \quad \lim\limits_{R\rightarrow +\infty }\left( \frac{1}{R^{2}}+ \frac{1}{R}+\zeta _{4, \frac{R}{2}}+\beta _{4, R}\right) = 0, $

取充分小的$ \varepsilon = \varepsilon_{1} $和充分大的$ R = R_{1}\geq 2 $使得

$ 2\sqrt{\left( \frac{1}{\sqrt{{\rho }}}+\frac{1}{\sqrt[4]{{\rho }}\sqrt{\beta }}\right) J_{\varepsilon _{1}, R_{1}}}\leq \frac{\eta }{2}. $

对固定的$ R_{1} $,由$ \mu _{m+1, R_{1}}\rightarrow \infty $$ (m\rightarrow \infty ) $知,存在充分大的$ m = m_{1} $使得

$ 2\sqrt{\frac{\beta +\frac{\alpha \beta ^{2}}{\sigma ^{2}}}{\sqrt{2\lambda +\mu _{m_{1}+1, R_{1}}}}}\leq \frac{\eta }{2}. $

因此

$ \widehat{\delta }_{\varepsilon _{1}, R_{1}, m_{1}} = 2\sqrt{\left( \frac{1}{ \sqrt{{\rho }}}+\frac{1}{\sqrt[4]{{\rho }}\sqrt{\beta }}\right) J_{\varepsilon _{1}, R_{1}}}+2\sqrt{\frac{\beta +\frac{\alpha \beta ^{2}}{\sigma ^{2}}}{\sqrt{2\lambda +\mu _{m_{1}+1, R_{1}}}}}\leq \eta . $

对任意$ D\in {\cal D}(E) $,令

$ \check{T}_{D}(\tau , \omega ) = {T}_{D}(\tau , \omega )+\max\{T_{\ast }(\omega), T_{\ast}(\theta_{-s} \omega), T_{\ast}(\theta_{-T_{\varepsilon_1}(\omega)}\omega)\}+ T_{\varepsilon_1}(\omega). $

根据以上讨论和性质(A),性质(B),引理3.6,引理3.8可知, $ \{{\chi }(\tau , \omega )\}_{\tau \in {{\Bbb R}} , \omega \in \Omega } $满足定理2.1中的(H1)–(H3).因此,定理3.1成立,且

$ l_{0} = \left[ \frac{4m_{1}\ln \left( \frac{2\sqrt{2m_{1}}}{\widehat{\delta }_{\varepsilon _{1}, R_{1}, m_{1}}}+1\right) }{\ln \frac{4}{3}}\right] +1, $

$ {\rm d}_{h}(\Psi (t, \tau -t, \theta _{-t}\omega )\, D(\tau -t, \theta _{-t}\omega ), \, \, {\cal O}(\tau , \omega ))\leq \check{b}_{\omega }e^{-\frac{ \ln \frac{4}{3}}{\max \{\frac{128\ln 2}{\rho }, \, \frac{16}{\sigma }\ln (1+ \frac{2\alpha \beta }{\sigma ^{2}})\}}t}, \quad t\geq \check{T}_{D}(\tau , \omega ), $

其中$ [\cdot ] $表示取整数.定理3.1证毕.

注3.1  定理3.1表明$ \Psi $存在随机吸引子$ {\cal \{A(\tau }, {\cal \omega )\}}_{\tau \in {{\Bbb R}} , \omega \in \Omega }\subseteq {\cal \{O(\tau }, {\cal \omega )\}} _{\tau \in {{\Bbb R}} , \omega \in \Omega } $,且其分形维数

$ \dim _{f}{\cal A}(\tau, \omega )\leq \dim _{f}{\cal O}(\tau , \omega )\leq l_{0}<\infty . $

因此,系统(1.1)的长期行为可以用有限个独立参数刻画.